A3 miner BCX
Publish: 2021-05-13 06:18:45
1. Don't bother. They can't compete with professional miners,
2. The following number is the drawing component number, and 2N3906 is its model. As for the replacement, the above friends have already explained, so it is unnecessary here. The symbol with * indicates that the component is very important in the circuit. It should be noted that it can not be replaced casually, and the parameters must be the same.
3. Unknown_Error
4. 1. What are the requirements for using "mobile Alibaba Wangwang"<
to use SMS Tradelink, you need to have the following conditions:
A. you must have a Chinese mobile phone
B. your mobile phone should have SMS service
C. you need to apply for an Aliwangwang account
D. please bind the mobile phone number with Tradelink
2. Mobile Alibaba Wangwang charges
binding mobile phone is free
sending SMS from Aliwangwang: the sender pays 0.1 yuan for each SMS, and the service provider dects the fee from Aliwangwang's account. The recipient is free of charge
reply SMS from mobile phone: Each SMS reply is 0.1 yuan, which is the same as ordinary SMS, and the operator (China Mobile or China Unicom) will dect the fee from the mobile phone account
both parties have no monthly rent
when Aliwangwang is offline, as long as you have bound your mobile phone and do not need to recharge, you can display Aliwangwang online (i.e. mobile phone online status)< br />
http://static.alisoft.com.cn/help/im//xd3xc3xbbxa7xcaxd6xb2xe1//.htm
to use SMS Tradelink, you need to have the following conditions:
A. you must have a Chinese mobile phone
B. your mobile phone should have SMS service
C. you need to apply for an Aliwangwang account
D. please bind the mobile phone number with Tradelink
2. Mobile Alibaba Wangwang charges
binding mobile phone is free
sending SMS from Aliwangwang: the sender pays 0.1 yuan for each SMS, and the service provider dects the fee from Aliwangwang's account. The recipient is free of charge
reply SMS from mobile phone: Each SMS reply is 0.1 yuan, which is the same as ordinary SMS, and the operator (China Mobile or China Unicom) will dect the fee from the mobile phone account
both parties have no monthly rent
when Aliwangwang is offline, as long as you have bound your mobile phone and do not need to recharge, you can display Aliwangwang online (i.e. mobile phone online status)< br />
http://static.alisoft.com.cn/help/im//xd3xc3xbbxa7xcaxd6xb2xe1//.htm
5. One factorization factor
1.2x4y2-4x3y2 + 10xy4< br />
2. 5xn+1－15xn＋60xn--1
3.
4. (a + b) 2x2-2 (A2-B2) XY + (a-b) 2Y2
5. X4-1
6. - A2-B2 + 2Ab + 4 factorization
7.
8.
9.
10. A2 + B2 + C2 + 2Ab + 2BC + 2Ac
11. X2-2x-8
12.3x2 + 5x-2
13. (x + 1) (x + 2) (x + 3) (x + 4) + 1
14. (x2 + 3x + 2) (x2 + 7x + 12) - 120.
15
16. Factorize the polynomial 5x2 ― 6xy ― 8y2< Verification: 32000-4 × 31999＋10 × 31998 is divisible by 7
18. Set it as a positive integer, and 64n-7n can be divided by 57. Prove that it is a multiple of 57.
19. Prove that no matter what the value of X and Y is, the value of is always positive
20. Given x2 + y2-4x + 6y + 13 = 0, find the value of X and y
three evaluations< Given that a, B and C satisfy A-B = 8, AB + C2 + 16 = 0, find the value of a + B + C.
22. Given that x2 + 3x + 6 is a factor of the polynomial x4-6x3 + MX2 + NX + 36, try to determine the value of M, N and find other factors< 1. Solution: the original formula = 2xy2 & # 8226; 2; x3－2xy2• 2x2＋2xy2• 5y2
=2xy2 (x3－2x2＋5y2)
prompt: first, determine the common factor and find the greatest common divisor 2 of each coefficient; The lowest power XY2 of the same letter of each item is the common factor 2xy2. Then the common factor of each item is mentioned outside the brackets, and the polynomial is written as the proct of the factor
2. Hint: in the common factor, the lowest power of the same letter X is xn -- 1, when xn + 1 extracts xn -- 1, it is X2, and xn extracts xn -- 1, it is X
solution: original formula = 5 xn -- 1 & 8226; x2－5xn--1• 3x＋5xn--1• 12
= 5 xn -- 1 (x2-3x + 12)
3. Solution: original formula = 3A (B-1) (1-8a3)
= 3A (B-1) (1-2a) (1 + 2A + 4a2) *
hint: cubic difference formula: a3-b3 = (a-b) (A2 + AB + B2)
cubic sum formula: A3 + B3 = (a + b) (a2-ab + B2)
so, 1-8a3 = (1-2a) (1 + 2A + 4a2)
4. Solution: the original formula = [(a + b) x] 2-2 (a + b) (a-b) XY + [(a-b) y] 2
= (AX + BX ay + by) 2
indicates that (a + b) x and (a-b) y are regarded as a whole
5. Solution: the original formula = (x2 + 1) (x2-1)
= (x2 + 1) (x + 1) (x-1)
hint: many students will not decompose until (x2 + 1) (x2-1), and factorization must not decompose any more
6. Solution: the original formula = - (a2-2ab + b2-4)
= - (a-b + 2) (a-b-2)
hint: if the first term of a polynomial is negative, it is generally necessary to put forward a negative sign so that the coefficient of the first term in brackets is positive. But we can't put forward the negative sign first. We should analyze the whole question to prevent the mistakes such as - 9x2 + 4y2 = (- 3x) 2 - (2Y) 2 = (- 3x + 2Y) (- 3x-2y) = (3x-2y) (3x + 2Y)
7. Solution: the original formula = x4-x3 - (x-1)
= X3 (x-1) - (x-1)
= (x-1) (x3-1)
= (x-1) 2 (x2 + X + 1) *
hint: in general, the factorization of four or more items should be appropriate, otherwise it cannot be decomposed. In addition, the result of this problem can't be written as (x-1) (x-1) (x2 + X + 1). If it can be written in the form of power, it must be written in the form of power* The cubic difference formula is used, (x-1) (x-1) (x-1) (x2 + X + 1) < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < 8: solution: the original formula = Y2 [(x + y) 2-12 (x + y) 2-12 (x + y) 2-12 (x + y) 2-12 (x + y) 2-12 (x + Y-12 (x + y) 2-12 (x + Y-12 (x + X + X + y + Y-12 (x + X + Y-12 (x + X + y-6) 2-12 (x + 12-12-12 (x + y-6) (x + 12-12-12-12-12 (x + X + 12-12-12-12-12) (x + 12-12-12 (x + X + 12-12-12-12-12) (x + 12-12-12-12-12-12-12) (x + 12) (x + 12-12- (x + y) 2]
= (x + y) 2 (X-6 + X + y) (x-6-x-y)
= (x + y) 2 (2x + y-6) (- 6-y)
= (x + y) 2 (2x + y-6) (y + 6)
10 The original formula = (A2 + B2 + 2Ab) + 2BC + 2Ac + C2
= (a + b) 2 + 2 (a + b) C + C2 *
= (a + B + C) 2
suggests: * regarding (a + B) as a whole
11. Solution: the original formula = x2-2x + 1-1-8 *
= (x-1) 2-32
= (x-1 + 3) (x-1-3)
= (x + 2) (x-4)
hint: this problem uses the collocation method, adding a "1" to x2-2x and subtracting a "1" to form a complete square formula
12. Solution: original formula = 3 (x2 + x) - 2
= 3 (x2 + X +) - 2 *
= 3 (x +) 2-3 × - 2
= 3 (x +) 2 -
= 3 [(x +) 2 -]
= 3 (x +) (x +)
= 3 (x + 2) (x -)
= (x + 2) (3x-1)
hint: * this step is very important, and it is based on the structure of complete square formula. For any quadratic trinomial AX2 + BX + C (a ≠ 0), it can be assigned to a (x +) 2 +.
13. Solution: the original formula = [(x + 1) (x + 4)] [(x + 2) (x + 3)] + 1
= (x2 + 5x + 4) (x2 + 5x + 6) + 1
let x2 + 5x = a, then the original formula = (a + 4) (a + 6) + 1
= A2 + 10A + 25
= (a + 5) 2
= (x2 + 5x + 5)
hint: take x2 + 5x as a whole
14. Solve the original formula = (x + 2) (x + 1) (x + 4) (x + 3) - 120
= (x + 2) (x + 3) (x + 1) (x + 4) - 120
= (x2 + 5x + 6) (x2 + 5x + 4) - 120
let x2 + 5x = m and substitute it into the above formula to get
the original formula = (M + 6) (M + 4) - 120 = M2 + 10m-96
= (M + 16) (M-6) = (x2 + 5x + 16) (x2 + 5x-6) = (x2 + 5x + 16) (x + 6) (x-1)
hint: take x2 + 5x as a whole
15. Solution: the original formula = (x + 2) (3x + 5)
prompt: decompose the quadratic term 3x2 into x and 3x (the quadratic term is usually only decomposed into positive factors), and the constant term 10 can be divided into 1 × 10＝－1 ×－ 10＝2 × 5＝－2 ×－ 5) Only 11x = x × 5+3x × 2
note: cross multiplication is a common method of factoring quadratic trinomial, especially when the coefficient of quadratic term is not 1, it brings us trouble in factoring. Here we mainly talk about this kind of situation. In decomposition, the quadratic term and the constant term are decomposed into the proct of two numbers respectively, and the proct of their cross multiplication is equal to the first term. It should be noted that: (1) if the constant term is a positive number, it should be decomposed into two factors with the same sign; if the primary term is positive, it is the same sign; If the primary term is negative, it should have a negative sign. (2) if the constant term is negative, it should be decomposed into two factors with different signs. In the proct of cross multiplication, the sign of the greater absolute value is the same as that of the primary term (if the primary term is positive, the greater absolute value in the proct of cross multiplication is the positive sign; If the primary term is negative, then the proct obtained by cross multiplication has the largest absolute value (negative sign)
ax C
quadratic constant term
BX d
ADX + BCX = (AD + BC) x linear term
abx2 + (AD + BC) x + CD = (AX + C) (BX + D)
16. Solution: original formula = (x-2y) (5x + 4Y)
x-2y
5x 4Y
- 6xy
17. Proof: original formula = 31998 (32-4) × 3＋10)= 31998 × 7,
can be divided by 7< Proof:
= 8 (82n-7n) + 8 × (82n-7n-7n-7n-7n-7n-7n + 7n + 7n + 7n + 7n (49 + 8)
= 8 (82n-7n-7n-7n) + 577nis the multiple of 57. < br /
< br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < 19.proof: the < br /
< br / < 19.proof: proof: the < br / < br / < br / < 19.proof: proof: the < br /
< br > < br / < br > < br > < br > < br > < br > < br > < br > < br / < br > < br > < br > < br > < br > < br > < br > < br / < br / < 20: 0
(X-2) 2 ≥ 0, (y + 3) 2 ≥ 0.
X-2 = 0 and y + 3 = 0
x = 2, y = - 3
triple evaluation
21. Solution: ∵ A-B = 8
A = 8 + B
and ab + C2 + 16 = 0
that is, (B + 8) B + C2 + 16 = 0
that is, (B + 4) 2 + C2 = 0
because, (B + 4) 2 ≥ 0, C2 ≥ 0,
b + 4 = 0, C = 0,
b = - 4, C = 0, a = B + 8 = 4
A + B + C = 0, Then
x4-6x3 + MX2 + NX + 36
= (x2 + PX + 6) (x2 + 3x + 6)
= X4 + (P + 3) X3 + (3P + 12) x2 + (6p + 18) x + 36
comparing the coefficients of both sides, the following equations are obtained:
the solution is obtained
1.2x4y2-4x3y2 + 10xy4< br />
2. 5xn+1－15xn＋60xn--1
3.
4. (a + b) 2x2-2 (A2-B2) XY + (a-b) 2Y2
5. X4-1
6. - A2-B2 + 2Ab + 4 factorization
7.
8.
9.
10. A2 + B2 + C2 + 2Ab + 2BC + 2Ac
11. X2-2x-8
12.3x2 + 5x-2
13. (x + 1) (x + 2) (x + 3) (x + 4) + 1
14. (x2 + 3x + 2) (x2 + 7x + 12) - 120.
15
16. Factorize the polynomial 5x2 ― 6xy ― 8y2< Verification: 32000-4 × 31999＋10 × 31998 is divisible by 7
18. Set it as a positive integer, and 64n-7n can be divided by 57. Prove that it is a multiple of 57.
19. Prove that no matter what the value of X and Y is, the value of is always positive
20. Given x2 + y2-4x + 6y + 13 = 0, find the value of X and y
three evaluations< Given that a, B and C satisfy A-B = 8, AB + C2 + 16 = 0, find the value of a + B + C.
22. Given that x2 + 3x + 6 is a factor of the polynomial x4-6x3 + MX2 + NX + 36, try to determine the value of M, N and find other factors< 1. Solution: the original formula = 2xy2 & # 8226; 2; x3－2xy2• 2x2＋2xy2• 5y2
=2xy2 (x3－2x2＋5y2)
prompt: first, determine the common factor and find the greatest common divisor 2 of each coefficient; The lowest power XY2 of the same letter of each item is the common factor 2xy2. Then the common factor of each item is mentioned outside the brackets, and the polynomial is written as the proct of the factor
2. Hint: in the common factor, the lowest power of the same letter X is xn -- 1, when xn + 1 extracts xn -- 1, it is X2, and xn extracts xn -- 1, it is X
solution: original formula = 5 xn -- 1 & 8226; x2－5xn--1• 3x＋5xn--1• 12
= 5 xn -- 1 (x2-3x + 12)
3. Solution: original formula = 3A (B-1) (1-8a3)
= 3A (B-1) (1-2a) (1 + 2A + 4a2) *
hint: cubic difference formula: a3-b3 = (a-b) (A2 + AB + B2)
cubic sum formula: A3 + B3 = (a + b) (a2-ab + B2)
so, 1-8a3 = (1-2a) (1 + 2A + 4a2)
4. Solution: the original formula = [(a + b) x] 2-2 (a + b) (a-b) XY + [(a-b) y] 2
= (AX + BX ay + by) 2
indicates that (a + b) x and (a-b) y are regarded as a whole
5. Solution: the original formula = (x2 + 1) (x2-1)
= (x2 + 1) (x + 1) (x-1)
hint: many students will not decompose until (x2 + 1) (x2-1), and factorization must not decompose any more
6. Solution: the original formula = - (a2-2ab + b2-4)
= - (a-b + 2) (a-b-2)
hint: if the first term of a polynomial is negative, it is generally necessary to put forward a negative sign so that the coefficient of the first term in brackets is positive. But we can't put forward the negative sign first. We should analyze the whole question to prevent the mistakes such as - 9x2 + 4y2 = (- 3x) 2 - (2Y) 2 = (- 3x + 2Y) (- 3x-2y) = (3x-2y) (3x + 2Y)
7. Solution: the original formula = x4-x3 - (x-1)
= X3 (x-1) - (x-1)
= (x-1) (x3-1)
= (x-1) 2 (x2 + X + 1) *
hint: in general, the factorization of four or more items should be appropriate, otherwise it cannot be decomposed. In addition, the result of this problem can't be written as (x-1) (x-1) (x2 + X + 1). If it can be written in the form of power, it must be written in the form of power* The cubic difference formula is used, (x-1) (x-1) (x-1) (x2 + X + 1) < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < 8: solution: the original formula = Y2 [(x + y) 2-12 (x + y) 2-12 (x + y) 2-12 (x + y) 2-12 (x + y) 2-12 (x + Y-12 (x + y) 2-12 (x + Y-12 (x + X + X + y + Y-12 (x + X + Y-12 (x + X + y-6) 2-12 (x + 12-12-12 (x + y-6) (x + 12-12-12-12-12 (x + X + 12-12-12-12-12) (x + 12-12-12 (x + X + 12-12-12-12-12) (x + 12-12-12-12-12-12-12) (x + 12) (x + 12-12- (x + y) 2]
= (x + y) 2 (X-6 + X + y) (x-6-x-y)
= (x + y) 2 (2x + y-6) (- 6-y)
= (x + y) 2 (2x + y-6) (y + 6)
10 The original formula = (A2 + B2 + 2Ab) + 2BC + 2Ac + C2
= (a + b) 2 + 2 (a + b) C + C2 *
= (a + B + C) 2
suggests: * regarding (a + B) as a whole
11. Solution: the original formula = x2-2x + 1-1-8 *
= (x-1) 2-32
= (x-1 + 3) (x-1-3)
= (x + 2) (x-4)
hint: this problem uses the collocation method, adding a "1" to x2-2x and subtracting a "1" to form a complete square formula
12. Solution: original formula = 3 (x2 + x) - 2
= 3 (x2 + X +) - 2 *
= 3 (x +) 2-3 × - 2
= 3 (x +) 2 -
= 3 [(x +) 2 -]
= 3 (x +) (x +)
= 3 (x + 2) (x -)
= (x + 2) (3x-1)
hint: * this step is very important, and it is based on the structure of complete square formula. For any quadratic trinomial AX2 + BX + C (a ≠ 0), it can be assigned to a (x +) 2 +.
13. Solution: the original formula = [(x + 1) (x + 4)] [(x + 2) (x + 3)] + 1
= (x2 + 5x + 4) (x2 + 5x + 6) + 1
let x2 + 5x = a, then the original formula = (a + 4) (a + 6) + 1
= A2 + 10A + 25
= (a + 5) 2
= (x2 + 5x + 5)
hint: take x2 + 5x as a whole
14. Solve the original formula = (x + 2) (x + 1) (x + 4) (x + 3) - 120
= (x + 2) (x + 3) (x + 1) (x + 4) - 120
= (x2 + 5x + 6) (x2 + 5x + 4) - 120
let x2 + 5x = m and substitute it into the above formula to get
the original formula = (M + 6) (M + 4) - 120 = M2 + 10m-96
= (M + 16) (M-6) = (x2 + 5x + 16) (x2 + 5x-6) = (x2 + 5x + 16) (x + 6) (x-1)
hint: take x2 + 5x as a whole
15. Solution: the original formula = (x + 2) (3x + 5)
prompt: decompose the quadratic term 3x2 into x and 3x (the quadratic term is usually only decomposed into positive factors), and the constant term 10 can be divided into 1 × 10＝－1 ×－ 10＝2 × 5＝－2 ×－ 5) Only 11x = x × 5+3x × 2
note: cross multiplication is a common method of factoring quadratic trinomial, especially when the coefficient of quadratic term is not 1, it brings us trouble in factoring. Here we mainly talk about this kind of situation. In decomposition, the quadratic term and the constant term are decomposed into the proct of two numbers respectively, and the proct of their cross multiplication is equal to the first term. It should be noted that: (1) if the constant term is a positive number, it should be decomposed into two factors with the same sign; if the primary term is positive, it is the same sign; If the primary term is negative, it should have a negative sign. (2) if the constant term is negative, it should be decomposed into two factors with different signs. In the proct of cross multiplication, the sign of the greater absolute value is the same as that of the primary term (if the primary term is positive, the greater absolute value in the proct of cross multiplication is the positive sign; If the primary term is negative, then the proct obtained by cross multiplication has the largest absolute value (negative sign)
ax C
quadratic constant term
BX d
ADX + BCX = (AD + BC) x linear term
abx2 + (AD + BC) x + CD = (AX + C) (BX + D)
16. Solution: original formula = (x-2y) (5x + 4Y)
x-2y
5x 4Y
- 6xy
17. Proof: original formula = 31998 (32-4) × 3＋10)= 31998 × 7,
can be divided by 7< Proof:
= 8 (82n-7n) + 8 × (82n-7n-7n-7n-7n-7n-7n + 7n + 7n + 7n + 7n (49 + 8)
= 8 (82n-7n-7n-7n) + 577nis the multiple of 57. < br /
< br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < br / < 19.proof: the < br /
< br / < 19.proof: proof: the < br / < br / < br / < 19.proof: proof: the < br /
< br > < br / < br > < br > < br > < br > < br > < br > < br > < br / < br > < br > < br > < br > < br > < br > < br > < br / < br / < 20: 0
(X-2) 2 ≥ 0, (y + 3) 2 ≥ 0.
X-2 = 0 and y + 3 = 0
x = 2, y = - 3
triple evaluation
21. Solution: ∵ A-B = 8
A = 8 + B
and ab + C2 + 16 = 0
that is, (B + 8) B + C2 + 16 = 0
that is, (B + 4) 2 + C2 = 0
because, (B + 4) 2 ≥ 0, C2 ≥ 0,
b + 4 = 0, C = 0,
b = - 4, C = 0, a = B + 8 = 4
A + B + C = 0, Then
x4-6x3 + MX2 + NX + 36
= (x2 + PX + 6) (x2 + 3x + 6)
= X4 + (P + 3) X3 + (3P + 12) x2 + (6p + 18) x + 36
comparing the coefficients of both sides, the following equations are obtained:
the solution is obtained
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