Question

1)(x+1)(x+2)=4(x+1) и x+2=4 2)x^2-1/x-1=5 и x+1=5 3)(x+1)^3=8 и x+1=2 4)x^2-1/x-1=2 и x+1=2

Answer 1

1)x^2+2x+x+2=4x+4
X^2-1x-2=0
D=1+8=9
X1=2
X2=-1
и
x+2=4
X=4-2=2

Answer 2

1. Let's solve the equation (x+1)(x+2) = 4(x+1):

Expanding the left side: x^2 + 3x + 2 = 4x + 4

Bringing all terms to one side: x^2 - x - 2 = 0

Factoring the quadratic equation: (x - 2)(x + 1) = 0

Setting each factor equal to zero: x - 2 = 0 or x + 1 = 0

Solving for x: x = 2 or x = -1

Therefore, the solutions to the equation are x = 2 and x = -1.

2. Let's solve the equation (x^2 - 1)/(x - 1) = 5:

Multiplying both sides of the equation by (x - 1) to eliminate the denominator: (x^2 - 1) = 5(x - 1)

Expanding: x^2 - 1 = 5x - 5

Bringing all terms to one side: x^2 - 5x + 4 = 0

Factoring the quadratic equation: (x - 4)(x - 1) = 0

Setting each factor equal to zero: x - 4 = 0 or x - 1 = 0

Solving for x: x = 4 or x = 1

Therefore, the solutions to the equation are x = 4 and x = 1.

3. Let's solve the equation (x+1)^3 = 8:

Taking the cube root of both sides: x + 1 = 2

Solving for x: x = 2 - 1 x = 1

Therefore, the solution to the equation is x = 1.

4. Let's solve the equation (x^2 - 1)/(x - 1) = 2:

Multiplying both sides of the equation by (x - 1) to eliminate the denominator: (x^2 - 1) = 2(x - 1)

Expanding: x^2 - 1 = 2x - 2

Bringing all terms to one side: x^2 - 2x - 1 = 0

This quadratic equation does not factor easily, so we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For this equation, a = 1, b = -2, and c = -1.

Substituting these values into the quadratic formula: x = (-(-2) ± √((-2)^2 - 4(1)(-1))) / (2(1)) x = (2 ± √(4 + 4)) / 2 x = (2 ± √8) / 2 x = (2 ± 2√2) / 2 x = 1 ± √2

Therefore, the solutions to the equation are x = 1 + √2 and x = 1 - √2.