Question

1)√х-1=х+3 2)√х+4=√х-6 3)³√х+2=х-1

Answer 1

1) \/x - 1=x +3 или \/x=x+4

Возведём обе части в квадрат

x=x²+8x+16 или

x²+7x+16=0

D=\/(49-64)=\/-15

Нет корней

2) \/x+4=\/x-6 откуда 4=-6

Нет корней

3) 3 3

\/x +2=x-1 или \/x=x-3

Возведём в куб обе части

x=x³-9x²+27x-27 откуда

x³-9x²+27x-27=0

x~=11.472

Answer 2

Let's solve each equation one by one:

1. √х - 1 = х + 3

To solve this equation, we need to isolate the square root term on one side of the equation. Follow these steps:

Step 1: Add 1 to both sides to eliminate the -1 on the left side: √х = х + 4

Step 2: Square both sides to get rid of the square root: (√х)² = (х + 4)² х = (х + 4)²

Step 3: Expand the right side: х = х² + 8х + 16

Step 4: Move all terms to one side to get a quadratic equation: х² + 8х + 16 - х = 0

Step 5: Combine like terms and set the equation to zero: х² + 7х + 16 = 0

This is a quadratic equation. To solve it, you can use the quadratic formula:

х = (-b ± √(b² - 4ac)) / 2a

where a = 1, b = 7, and c = 16.

х = (-7 ± √(7² - 4 * 1 * 16)) / 2 * 1 х = (-7 ± √(49 - 64)) / 2 х = (-7 ± √(-15)) / 2

Since the square root of a negative number is not a real number, there are no real solutions for this equation.

2. √х + 4 = √х - 6

To solve this equation, we'll square both sides to eliminate the square roots:

(√х + 4)² = (√х - 6)²

х + 8√х + 16 = х - 12√х + 36

Now, let's move all the terms containing √х to one side and the constant terms to the other side:

8√х + 12√х = 36 - 16

20√х = 20

Now, divide both sides by 20:

√х = 1

Square both sides to find х:

(√х)² = 1²

х = 1

So, the solution to the equation is х = 1.

3. ³√х + 2 = х - 1

To solve this equation, we need to isolate the cube root term on one side of the equation:

³√х = х - 1 - 2

³√х = х - 3

Now, cube both sides to eliminate the cube root:

(³√х)³ = (х - 3)³

х = (х - 3)³

Expand the right side:

х = х³ - 9х² + 27х - 27

Move all terms to one side to get a cubic equation:

х³ - 9х² + 27х - 27 - х = 0

Simplify:

х³ - 9х² + 26х - 27 = 0

Unfortunately, solving cubic equations is generally more complex than solving quadratics, and there is no simple formula like the quadratic formula for cubic equations. In this case, you might need to use numerical methods or graphical methods to approximate the solution. If you have specific values for x that you're interested in, you can plug them into the equation to find the corresponding y-values.