If the expansion of (x ^ 2 + mx-8) (x ^ 2-3x + n) does not contain x ^ 2 and x ^ 3 terms, find the values of M and n

If the expansion of (x ^ 2 + mx-8) (x ^ 2-3x + n) does not contain x ^ 2 and x ^ 3 terms, find the values of M and n

Just expand the items with X & # 178; and X & # 179?
The expansion is NX & # 178; - 3mx & # 178; - 8x & # 178; - 3x & # 179; + MX & # 179;
From the meaning of the title, n-3m-8 = 0
-3+m=0
The solution is m = 3, n = 17

If (x-3) (3x + 2) = 3x ^ 2 + MX + N, then M=__ n=___ If (X-8) (x ^ 2-x + m) does not contain a linear term, then the value of M is

(x-3) (3x-3) (3x-3) (3x-3) (3x-3) (3x-3) (3x-3) (3x-3) (3x-3) (3x-3) (3x-3) (3x-3) (3x-2x-9x-6 = 3x \35;#178; (2x-9x-6; (2x-9x-6 = 3x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\it's not easy

If there is no quadratic term of X in the product of (x2-nx + 3) (3x-2), then n=______ ．

The original formula = 3x3-2x2-3nx2 + 9x-6, = 3x3 - (2 + 3n) x2 + 9x-6, ∵ there is no quadratic term of X in the product, ∵ (2 + 3n) = 0, ∵ n = - 23, so the answer is: - 23

If X1 and X2 are two real roots of the equation x ^ 2 - (2k + 1) x + K ^ 2 + 1 = 0 about X,
If x 1 and x 2 are two real number roots of the equation x ^ 2 - (2k + 1) x + K ^ 2 + 1 = 0, and x 1 and x 2 are both greater than 1, the value range of real number k is obtained
We can know that x 1 + x 2 = 2K + 1 is greater than 2, we can get k greater than 1 / 2, and x 1 * x 2 = k ^ 2 + 1 is greater than 1. But my answer is k ≥ 3 / 4 and K is not equal to 1. What's the matter? What's wrong with my operation? Please explain in detail

Note that this kind of problem is wrong with Veda's theorem! It should be done with the distribution of roots!
The symmetry axis X = K + 1 / 2
Then three conditions can be listed: 1, f (1) > 0.2, x = K + 1 / 2 > 1.3, △ ≥ 0
Then work out these three conditions, and it's OK

Urgent! To solve binary linear equations: {- 2x + 3Y = 19,6x + 7Y = - 9!

If we multiply Formula 1 by 3 and add formula 2, we can get 16y = 48, y = 3. If we take Formula 1, we can get x = - 5,

8:7 plus 2x equals 13:5. (solving equation) although it's an oral arithmetic problem, I didn't listen to it in class and can't do it

1)2x+8=16(2)x/5=10(3)x+7x=8(4)9x-3x=6(5)6x-8=4(6)5x+x=9(7)x-8=6x(8)4/5x=20(9)2x-6=12(10)7x+7=14(11)6x-6=0(12)5x+6=11(13)2x-8=10(14)1/2x-8=4(15)x-5/6=7(16)3x+7=28(17)3x-7=26(18)9x-x=16(19)24x+x=50(20)6...

When Xiao Ming solved the equation 2x − 15 + 1 = x + A2, due to carelessness, when the denominator is removed, the left 1 of the equation is not multiplied by 10, so the solution obtained is x = 4. Try to find the value of a, and correctly find the solution of the equation

When the denominator is removed, only 1 on the left side of the equation is not multiplied by 10, 2 (2x-1) + 1 = 5 (x + a). Substituting x = 4 into the above formula, the solution is a = - 1. The original equation can be reduced to: 2x − 15 + 1 = x − 12, the denominator is removed, 2 (2x-1) + 10 = 5 (x-1) is obtained, the brackets are removed, 4x-2 + 10 = 5x-5 is obtained, the terms of the same kind are combined, and the - x = - 13 system is obtained

When Xiao Ming solved the equation 2x-3 / 5 + 1 = x + A / 2, due to carelessness, when the denominator was removed, the left 1 of the equation was not multiplied by 10, so the solution was x = 4. Try to find the value of a, and correctly find the solution of the equation

If it is not multiplied by 10, then x = 4 is the solution of the equation 20x-6 + 1 = 10x + 5a, that is, 10x-5-5a = 0. Substituting x = 4 into the new equation, 40-5-5a = 0 is obtained, and a = 7 is calculated. Then the original equation is 2x-3 / 5 + 1 = x + 14, x = 68 / 5

When Xiao Ming solved the equation 2x − 15 + 1 = x + A2, due to carelessness, when the denominator is removed, the left 1 of the equation is not multiplied by 10, so the solution obtained is x = 4. Try to find the value of a, and correctly find the solution of the equation

∵ when the denominator is removed, only 1 on the left side of the equation is not multiplied by 10, 2 (2x-1) + 1 = 5 (x + a), substituting x = 4 into the above formula, the solution is a = - 1. The original equation can be reduced to: 2x − 15 + 1 = x − 12, the denominator is removed, 2 (2x-1) + 10 = 5 (x-1) is removed, the bracket is removed, 4x-2 + 10 = 5x-5 is obtained, the term is shifted and the similar term is merged, the - x = - 13 coefficient is changed to 1, x = 13, so a = - 1, x = 13

When Xiao Ming solved the equation 2x − 15 + 1 = x + A2, due to carelessness, when the denominator is removed, the left 1 of the equation is not multiplied by 10, so the solution obtained is x = 4. Try to find the value of a, and correctly find the solution of the equation

When the denominator is removed, only 1 on the left side of the equation is not multiplied by 10, 2 (2x-1) + 1 = 5 (x + a). Substituting x = 4 into the above formula, the solution is a = - 1. The original equation can be reduced to: 2x − 15 + 1 = x − 12, the denominator is removed, 2 (2x-1) + 10 = 5 (x-1) is obtained, the brackets are removed, 4x-2 + 10 = 5x-5 is obtained, the terms of the same kind are combined, and the - x = - 13 system is obtained