The solution equation is: (1) x ^ 2 + x = 0 (2) (X-2) (x-3) = 30

The solution equation is: (1) x ^ 2 + x = 0 (2) (X-2) (x-3) = 30

（1）x^2+x=0
x（x+1）=0
X = 0 or x = - 1
（2） （x-2）（x-3）=30
Expansion, X & # 178; - 5x + 6 = 30
x²-5x-24=0
（x+3)（x-8）=0
X = - 3 or x = 8

Prove: the equation AX & # 178; + BX + C = 0 (a ≠ 0) about X, if a + B + C = 0, then its two are X1 = 1, X2 = (- a-b) / a respectively

On the equation AX & # 178; + BX + C = 0 (a ≠ 0), if a + B + C = 0
Obviously, there is a root x = 1
Because substituting x = 1 gives a + B + C = 0
And then the sum of the two
=-b/a
So the other one is
-b/a-1=（-a-b）/a

Solve equation 30 / 30 + x = 0.4
process

Multiply both sides by 30 + X
30=0.4(30+x)
30=12+0.4x
0.4x=30-12=18
x=18÷0.4
x=45

It is known that the quadratic trinomial x ^ 2 + MX + n of X has a factor (x + 5), and M + n = 17. Try to find the value of M, n
After factoring the factor (x ^ 2 + 5x + 3) (x ^ 2 + 5x-23) + k = (x ^ 2 + 5x-10) ^ 2, find the value of K
Let a = (a ^ 2 + 1) (b ^ 2 + 1) - 4AB
(1) Try to write the polynomial as the sum of two nonnegative numbers
(2) Let a = 0, find the value of a, B

1. If there is a factor (x + 5), then a solution of the equation x ^ 2 + MX + n = 0 is - 5
Then 25-5m + n = 0
m+n=17
So m = 7, n = 10
2. Let y = x ^ 2 + 5x
Then (y + 3) (y-23) + k = (Y-10) ^ 2
y^2-20y-69+k=y^2-20y+100
k=169
3.A=a^2b^2+a^2+b^2+1-4ab=a^2b^2-2ab+1+a^2-2ab+b^2=(ab-1)^2+(a-b)^2
A=0 ab-1=0 a-b=0
A = b = 1 or a = b = - 1

The solution equation is: (3 + x) / (30 + x) = 0.25

Let's take this as a fraction and multiply it diagonally
We get 12 + 4x = 30 + X
x=6

It is known that there is a factor (x + 5) for the quadratic trinomial X & # 178; + MX + n of X, and M + n = 7?

（x+5）（x+k）=x²+（5+k)x+5k
m=5+k
n=5k
That is, n = 5 (m-5) = 5m-25
m+n=7
M = 16 / 3
n=5/3

If x-3 of a plus x + 4 of B equals 2x + 1 (x-3) multiplied by (x + 4), then a equals? B equals?

Please confirm the title! A / (x-3) + B / (x + 4) = (2x + 1) / [(x-3) (x + 4) [a (x + 4) + B (x-3)] / [(x-3) (x + 4]) = (2x + 1) / [(x-3) (x + 4) ∧ a (x + 4) + B (x-3) = 2x + 1 ∧ (a + b) x + 4a-3b = 2x + 1 no matter what the value of X is, the equation holds ∧ a + B = 2 and 4a-3b = 1, the solution is: a = 1, B = 1

It is known that there is a factor (x + 5) for the square + MX + n of the quadratic trinomial X of X. m + n = 17. Try to find the value of M.N

It is known that the quadratic trinomial can be reduced to (x + 5) (x +?). When x = 1, there is 1 + m + n = 18, so there is 6 * (1 +?) = 18, so? = 2,
So if we take (x + 5) (x + 2), we can get m = 7, n = 10, where we can set the unknown y,
The second solution, from known, quadratic trinomial can be changed into (x + 5) (x + y), from known 5Y = n, 5 + y = m,
M + n = 17, so 5Y + 5 + y = 17, y = 2, so m = 7, n = 10

5×（3-2x）=2.4×56

5×（3-2x）=2.4×56
15-15x=140
-15x=125
x=-8.333

It is known that there is a factor (x + 3) in the quadratic trinomial X & # 178; - 4mx + M. find another factor and the value of M

Substitute x = - 3 into the equation x & # 178; - 4mx + M = 0
9+12m+m=0
m=-9/13
Because a factor is: x-3 / 13