If x + 2Y + 3Z = 11,3x + 5Y + 7z = 27, find the value of X + y + Z RT

If x + 2Y + 3Z = 11,3x + 5Y + 7z = 27, find the value of X + y + Z RT

3x + 5Y + 7z = 27 minus x + 2Y + 3Z = 11
We get 2x + 3Y + 4Z = 16 and then subtract x + 2Y + 3Z = 11
x+y+z=5

Given the equations 3x + 5Y + 3Z = 0, 3x-5y-8z = 0, and Z ≠ 0, find X: Z and Y: Z

By adding the two formulas, 6x-5z = 0, that is, x = 5Z / 6, that is, X / z = 5 / 6
Then, substituting x = 5Z / 6 into Formula 1, we get 5Y + 11z / 2 = 0 and Y / z = - 11 / 10

The root of quadratic equation x ^ 2-2x-1 = 0 is ()

The discriminant of root △ = (- 2) & sup2; - 4 × 1 × (- 1) = 8 > 0
So there are two different roots

On the quadratic equation K (X & sup2; - 2x + 1) - 2x & sup2; + x = 0 with one variable of X, there are two real roots, so we can find the value of K

k（x²-2x+1）-2x²+x=0
(k-2)x^2+(1-2k)x+k=0
If there are two, then K-2 ≠ 0, K ≠ 2
△=(1-2k)^2-4k(k-2)≥0
1-4k+4k^2-4k^2+8k≥0
4k≥-1
k≥-1/4
The range of K is k ≥ - 1 / 4 and K ≠ 2

The following is the case of judging the root of the known quadratic equation x & sup2; + √ 3kx + K & sup2; - K + 2 = 0 with respect to X
Please judge whether the answer to this question is correct. If there is any mistake, please write down the correct answer
The solution: △ = B & sup2; - 4ac = (√ 3K) & sup2; - 4 (K & sup2; - K + 2) = (K-2) & sup2; + 4,
∵（k-2）²≥0,∴（k-2）²+＞0,
The original equation has two unequal real roots

Calculation error of △ in the process of solving!
△=b²-4ac=（√3k）²-4（k²-k+2）=-k²+4k-8=-（k-2）²-4
∵（k-2）²≥0,∴-（k-2）²-4

We know the quadratic equation x2 + K (x-1) - 1 = 0 (1) about X. we prove that no matter what the value of K is, the equation always has two real roots; (2) is there a positive number k, so that the two real roots X1 and X2 of the equation satisfy x12 + kx1 + 2x1x2 = 7-3 (x1 + x2)? If it exists, try to find the value of K; if it does not exist, explain the reason

(1) The equation x2 + K (x-1) - 1 = 0 can be reduced to x2 + kx-k-1 = 0, because △ = K2 + 4K + 4 = (K + 2) 2 ≥ 0, so the equation has two real roots. (2) suppose that there is a positive number k, satisfying X12 + kx1 + 2x1x2 = 7-3 (x1 + x2), since X1 and X2 are the two real roots of the equation, we substitute x = X1 to get: X12 + kx1-k-1 = 0, ｜ X12 + kx1 = K + 1, X1 + x2 = - K, x1x2 = - k-1, that is, K + 1 + 2 (- k-1) = 7 + 3k, and get k = - 2 This is in contradiction with the problem k > 0. The positive number k satisfying the condition does not exist

If (M + 2) x ^ 2 + x-m ^ 2-5m-6 = 0 has a root of 0, then M=____ The other root is_______ .

The equation (M + 2) x ^ 2 + x-m ^ 2-5m-6 = 0 has a root of 0
-m^2-5m-6=0
m=-2 m=-3
Because there's another root, M + 2 doesn't equal 0
So m = - 2 is left out
That is, M = - 3
Equation - x ^ 2 + x = 0
The other root is 1

For the quadratic equation (M + 2) x ^ 2 + x-m ^ 2-5m-6 = 0 with a root of 0, then M=

On X, the quadratic equation (M + 2) x ^ 2 + x-m ^ 2-5m-6 = 0 has a root of 0
The results show that - M & # 178; - 5m-6 = 0
m²+5m+6=0
(m+2)(m+3)=0
m1=-2,m2=-3
And ∵ m + 2 ≠ 0
Ψ take M = - 3

Solving quadratic equation of one variable: X (5-x) = 7

x（5-x）=7
x²-5x+7 = 0
△= 25 - 28 = -3 < 0
No real solution

1. If the quadratic coefficient of a quadratic equation of one variable is 1 and its two roots are 1, - 2, then the equation___

(x-1)(x+2）=0
∴x²+x-2=0