It is known that the solution of equation 6 (X-2) = 5x is twice the solution of equation 2 (2x-3) - 3 (1-A) = 2x about X, and the value of a is obtained

It is known that the solution of equation 6 (X-2) = 5x is twice the solution of equation 2 (2x-3) - 3 (1-A) = 2x about X, and the value of a is obtained

6(x-2)=5x
6x-12=5x
x=12
Then the solution of 2 (2x-3) - 3 (1-A) = 2x is 12 / 2 = 6
Substituting x = 6,
18-3(1-a)=12
1-a=2
a=-1

How to calculate the equation {5x + 4Y = - 3,2x-3y = 8}

{5x+4y=-3 1)
2x-3y=8 2)
1 × 2-2) × 5
23y=-46
y=-2
Substituting 2), we get
x=1

The solution of equation 6x2 + 5x − 32x + 10 = 0 is______ ．

The two sides of the equation are multiplied by 2x (x + 5), the denominator is removed to get 6 × 2-3x = 0, and the solution is x = 4. After testing, x = 4 is the solution of the original equation. Therefore, the answer to this question is: x = 4

Offering reward for 200 points to solve the exponential equation 3 * 2 ^ 2x + 2 * 3 ^ 2x-5 * 6 ^ x = 0
I hope I can give the process. I find that x = 0 or 1 has to be proved, but I don't know how to prove it
△=25b^2-24b^2=b^2
A = (5b + b) / 6 = B, or a = (5b-b) / 6 = 2B / 3
How did this come out

3*2^2x+2*3^2x-5*6^x=0
2^x=a,3^x=b,6^x=ab
3*2^2x+2*3^2x-5*6^x=0
3a^2+2b^2-5ab=0
△=25b^2-24b^2=b^2
A = (5b + b) / 6 = B, or a = (5b-b) / 6 = 2B / 3
A = B or a = 2 / 3B
1. When a = b
2^x=3^x
x=0
2. When a = 2 / 3B
2^x=2/3*3^x
2^(x-1)=3^(x-1)
(x-1) = (x-1) log2 (3) log2 (3) with 2 as the base and 3 as the index
So X-1 = 0, x = 1
So, x = 1, or x = 0

Write out the solution set of the following equation (1) 2x-1 = 0 (2) 4 (x + 1) - 3 (x + 1) = 2 (3) x to the second power - 5x + 4 = 0 (4) x2 + X-1 = 0

1) By solving the equation 2x-1 = 0, we get x = 1 / 2, and the solution set of the equation is {1 / 2};
2) By solving the equation 4 (x + 1) - 3 (x + 1) = 2, we get x = 1, and the solution set of the equation is {1};
3) By solving the equation x & sup2; - 5x + 4 = 0, we obtain x = 1 or x = 4, and the solution set of the equation is {1,4};
4) By solving the equation x & sup2; + X-1 = 0, we get x = (- 1 ± √ 5) / 2, and the solution set of the equation is {(- 1 ＋ √ 5) / 2, (- 1 - √ 5) / 2}

The third power of 15x-the second power of 40x and the second power of y-5x

The third power of 15x-the second power of 40x and the second power of y-5x
=15x³-40x²y-5x²
=5x²(3x-8y-1)

The square of (5x-1)

（5x-1）²
=25x²-10x+1
 

Square of x-5x + 4 = (square of x-1) - (5x -?)

Square of X - 5x + 4 = (square of X - 1) - (5x - 5)

In the case of the square of X + 5x-1 = 0 root

△=b^2-4ac=5^2-4*1*(-1)=25+4=29＞0
The equation has two unequal real roots

0.5x + 2 / 5 = 80% - 0.08

0.5x + 2 / 5 = 80% - 0.080.5x = 0.8-0.08-0.40.5x = 0.32x = 0.64
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