Question

1.(2x+5)(3x-7)=(1-2x)(1+2x)+(5x+)(2x-5) 2.(2x-1)(2x+1)-4(x+3)^2=71 помогите пожалуйста ,умоляю)

Answer 1

1)(2х+5)(3х-7)=(1-2х)(1+2х)+(5х+3 )(2х-5)
6х²-14х+15х-35=1-4х²+10х²-25х+6х-15
6х²-14х+15х-35-1+4х²-10х²+25х-6х+15=0
20х=-11
х=-11/20=-0,55
2)(2х-1)(2х+1)-4(х+3)²=71
4х²-1-4х²-24х-36=71
-24х=71+1+36
-24х=108
х=108:(-24)
х=-4,5

Answer 2

1. Solving the equation (2x+5)(3x-7)=(1-2x)(1+2x)+(5x+)(2x-5)

To solve the equation (2x+5)(3x-7)=(1-2x)(1+2x)+(5x+)(2x-5), we need to simplify both sides of the equation and then solve for x.

Let's start by simplifying the left side of the equation: (2x+5)(3x-7) = 6x^2 - 14x + 15x - 35 = 6x^2 + x - 35

Now, let's simplify the right side of the equation: (1-2x)(1+2x) = 1 - 2x + 2x - 4x^2 = 1 - 4x^2 (5x+)(2x-5) = 5x(2x-5) = 10x^2 - 25x

Combining the simplified right side of the equation: (1-4x^2) + (10x^2 - 25x) = 1 - 4x^2 + 10x^2 - 25x = 6x^2 - 25x + 1

Now, we have the equation: 6x^2 + x - 35 = 6x^2 - 25x + 1

To solve for x, we can subtract 6x^2 from both sides of the equation: x - 35 = -25x + 1

Next, let's add 25x to both sides of the equation: 26x - 35 = 1

Finally, we can add 35 to both sides of the equation: 26x = 36

Dividing both sides of the equation by 26, we find: x = 36/26 = 18/13

Therefore, the solution to the equation (2x+5)(3x-7)=(1-2x)(1+2x)+(5x+)(2x-5) is x = 18/13.

2. Solving the equation (2x-1)(2x+1)-4(x+3)^2=71

To solve the equation (2x-1)(2x+1)-4(x+3)^2=71, we need to simplify both sides of the equation and then solve for x.

Let's start by simplifying the left side of the equation: (2x-1)(2x+1) = 4x^2 - 1 -4(x+3)^2 = -4(x^2 + 6x + 9) = -4x^2 - 24x - 36

Now, we have the equation: 4x^2 - 1 - 4x^2 - 24x - 36 = 71

Simplifying further: -1 - 24x - 36 = 71 -60 - 24x = 71

To solve for x, we can add 60 to both sides of the equation: -24x = 131

Finally, we can divide both sides of the equation by -24: x = 131/-24

Therefore, the solution to the equation (2x-1)(2x+1)-4(x+3)^2=71 is x = 131/-24.

Please note that the solutions provided are based on the given equations and calculations.