Question

1) (3-2)^2=48+(2-х)^2 2) (1+х)^2=(1-2х)^2 3) х^2+5х+42=х(2х+6)

Answer 1

1
48+(2-x)²=1
(2-x)²=-47
нет решения
2
(1+x)²-(1-2x)²=0
(1+x-1+2x)(1+x+1-2x)=0
3x*(2-x)=0
x=0  x=2
3
x²+5x+42-2x²-6x=0
-x²-x+42=0
x²+x-42=0
x1+x2=-1 U x1*x2=-42
x1=-7 U x2=6

Answer 2

Let's solve each of the equations step by step:

1. (3-2)^2 = 48 + (2-x)^2

Simplifying the left side: 1^2 = 1

Expanding the right side: 48 + (2-x)^2 = 48 + (2-x)(2-x) = 48 + 4 - 4x - 4x + x^2 = 52 - 8x + x^2

Now we can set up the equation: 1 = 52 - 8x + x^2

Rearranging the terms: x^2 - 8x + 51 = 0

This is a quadratic equation. We can either factor it or use the quadratic formula to solve for x. In this case, the equation does not factor nicely, so let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 1, b = -8, and c = 51:

x = (-(-8) ± √((-8)^2 - 4(1)(51))) / (2(1)) x = (8 ± √(64 - 204)) / 2 x = (8 ± √(-140)) / 2

Since the discriminant is negative, the equation has no real solutions. Therefore, there is no real value of x that satisfies this equation.

2. (1+x)^2 = (1-2x)^2

Expanding both sides: (1+x)(1+x) = (1-2x)(1-2x) 1 + 2x + x^2 = 1 - 4x + 4x^2

Rearranging the terms: 0 = 3x^2 - 6x

Factoring out x: 0 = 3x(x - 2)

Setting each factor equal to zero: x = 0 or x - 2 = 0 x = 0 or x = 2

So, the equation has two solutions: x = 0 and x = 2.

3. x^2 + 5x + 42 = x(2x + 6)

Expanding the right side: x^2 + 5x + 42 = 2x^2 + 6x

Rearranging the terms: 0 = 2x^2 + 6x - x^2 - 5x - 42 0 = x^2 + x - 42

Factoring the quadratic equation: 0 = (x - 6)(x + 7)

Setting each factor equal to zero: x - 6 = 0 or x + 7 = 0 x = 6 or x = -7

So, the equation has two solutions: x = 6 and x = -7.