Question

1)y = (x + 1)^2, y = 1; 2)y = x² +1, y = 5;3)y= -x^2 + 4, y = 0;4)y=x^3, y=x, x=0 x=15)y=3-x^2, y=26)x^3+1, y=1, x=1​

Answer 1

Ответ:

ЗАДАНИЕ 1.

1)y = (x + 1)², y=1

(x+1)² =1

x+1= -1 или  x+1= 1

x= - 2;           x= 0

Ответ: x=-2; x=0

ЗАДАНИЕ 2.

2)y = x² +1, y = 5;

x²+1=5

x² =4

x=2 или x=-2

Ответ: x=2; x=-2

ЗАДАНИЕ 3.

3)y= -x² + 4, y = 0

-x² +4=0

-x²=-4

x²=4

x=2   или  x=-2

Ответ: x=2; x=-2

ЗАДАНИЕ 4.

y=x^3, y=x, x=0, x=1

*Если y=x, то:

x= x^3

x-x^3=0

x(1-x²)=0

x=0      или      1-x²=0

                   x=-1 или x=1

Ответ: x=0; x=-1; x=1.

*Если x=0, то:

y=0^3

y=0

Ответ: y=0

*Если x=1, то:

y= 1^3

y=1

Ответ: y=1

ЗАДАНИЕ 5.

y=3-x²,  y=2

3-x²=2

-x² =2-3

x²=1

x=1 или x=-1

Ответ: x=1; x=-1

ЗАДАНИЕ 6.

y=x^3+1, y=1, x=1

*Если y=1, то:

x^3+1=1

x^3=1-1

x^3=0

x=0

Ответ: x=0

*Если x=1, то

y=1^3+1

y=1+1

y=2

Ответ: y=2

Answer 2

Let's go through each of the given equations and find the corresponding values of x when y equals the given values.

1. y = (x + 1)^2, y = 1: To find the values of x when y = 1, we substitute y = 1 into the equation: 1 = (x + 1)^2

Taking the square root of both sides, we have: √1 = √((x + 1)^2) 1 = x + 1

Simplifying the equation, we get: x = 1 - 1 x = 0

So, when y = 1, x = 0.

2. y = x^2 + 1, y = 5: Substituting y = 5 into the equation: 5 = x^2 + 1

Rearranging the equation, we have: x^2 = 5 - 1 x^2 = 4

Taking the square root of both sides, we get: x = ±√4

Therefore, when y = 5, x can be either x = 2 or x = -2.

3. y = -x^2 + 4, y = 0: Substituting y = 0 into the equation: 0 = -x^2 + 4

Rearranging the equation, we have: x^2 = 4

Taking the square root of both sides, we get: x = ±√4

Therefore, when y = 0, x can be either x = 2 or x = -2.

4. y = x^3, y = x, x = 0, x = 1: Substituting x = 0 into the equation: y = 0^3 y = 0

Substituting x = 1 into the equation: y = 1^3 y = 1

So, when x = 0, y = 0, and when x = 1, y = 1.

5. y = 3 - x^2, y = 2: Substituting y = 2 into the equation: 2 = 3 - x^2

Rearranging the equation, we have: x^2 = 3 - 2 x^2 = 1

Taking the square root of both sides, we get: x = ±√1

Therefore, when y = 2, x can be either x = 1 or x = -1.

6. y = x^3 + 1, y = 1, x = 1: Substituting y = 1 and x = 1 into the equation: 1 = 1^3 + 1 1 = 1 + 1 1 = 2

The equation is not satisfied when y = 1 and x = 1.

So, there is no solution for this equation when y = 1 and x = 1.

Please let me know if you need any further assistance!