Algebraic evaluation exercises When A-B / A + B = 3, find the value of algebraic formula 2 (a-b) / (a + b) - 4 (a + b) / 3 (a-b)

Algebraic evaluation exercises When A-B / A + B = 3, find the value of algebraic formula 2 (a-b) / (a + b) - 4 (a + b) / 3 (a-b)

From (a-b) / (a + b) = 3, it is obtained that:
(a+b)/(a-b)=1/3.
By substituting (a-b) / (a + b) = 3 and (a + b) / (a-b) = 1 / 3, the following results are obtained:
2(a-b)/(a+b)-4(a+b)/3(a-b)
=2*3-4/3*1/3
=50/9

When m is a value, the quadratic equation MX & # 178; + MX = 2m-9 has two equal real roots

mx²+mx=2m—9
mx²+mx-(2m-9)=0
△=m²+4m(2m-9)
=m²+8m²-36m
=9m²-36m=0
M = 0 (rounding off)
m=4
∴m=4

The solution of inequality about X: MX & # 178; - 3 (M + 1) x + 9 > 0 (m ∈ R)
If the inequality X & # 178; + ax + 1 ≥ 0 holds for all x ∈ (0,1 / 2), find the value range of A

One question:
When m = 0, the solution of - 3x + 9 > 0 is x0
(1) When m

It is known that α and β are the two roots of the equation x & # 178; + (M + 2) x + 1 = 0. Find the value of (α & # 178; + m α + 1) (β & # 178; + m β + 1)

It is known that α and β are two roots of the equation x & # 178; + (M + 2) x + 1 = 0, then: α & # 178; + (M + 2) α + 1 = 0, α & # 178; + m α + 1 = - 2 α β & # 178; + (M + 2) β + 1 = 0, β & # 178; + m β + 1 = - 2 β, according to Weida's theorem: α β = 1 / 1 = 1; (α & # 178; + m α + 1) (β & # 178; + m β + 1) = - 2 α (- 2 β) =

If M / (x + 1) + n / (x-1) = (1-5x) / (x ^ 2-1) holds, find the value of M and n,

M (x-1) + n (x + 1) = 1-5x
(M+N)x+N-M=-5x+1
Because the above formula holds, so:
M+N=-5
N-M=1
N=-2 M=-3
Absolutely right,

What is the value of 1 m when (m-1) x ^ 2-2 (m-1) x + 3 (M + 1) < 0 holds for all x ∈ r?
2 when a > 1, the solution set of inequality (x-1) (x-a) < 0 is
3 when a ＜ - 1, the solution set of inequality a (x + a) (x-1) ＜ 0 is

1. It is required that (m-1) x ^ 2-2 (m-1) x + 3 (M + 1) < 0 (x ∈ R), M-1 is required

When k is the value, the sum of two squares of x ^ 2-kx-2 + k = 0 is the smallest

Sum of squares X1 ^ 2 + x2 ^ 2 = (x1 + x2) ^ 2-2x1x2
According to Weida's theorem, we get X1 + x2 = k, x1x2 = - 2 + K, so X1 ^ 2 + x2 ^ 2 = k ^ 2-2k + 4 = (k-1) ^ 2 + 3,
Because (k-1) ^ 2 > = 0, when k = 1, X1 ^ 2 + x2 ^ 2 gets the minimum value of 3

[y (X-Y) + X (x + y) - X & # 178;] / (- y) process

[y(x-y)+x(x+y)-x²]÷(-y)
=(xy-y^2+x^2+xy-x^2)÷(-y)
=(2xy-y^2)÷(-y)
=-2x+y