Реши уравнение:1)√x+√-x=0, 2)√x+√-x=1, 3)√x²-2x+1+√x²-1=0 4)(x-2)√x-3=0

1. To solve the equation √x + √(-x) = 0, we need to solve for x.

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

2. To solve the equation √x + √(-x) = 1, we need to solve for x.

Again, since the square root of a negative number is not a real number, there are no real solutions to this equation.

3. To solve the equation √x² - 2x + 1 + √x² - 1 = 0, we need to simplify it first.

√x² - 2x + 1 + √x² - 1 = 0 2√x² - 2x = 0 √x² - x = 0

Now, we can solve for x.

√x² - x = 0 x² - x = 0 x(x - 1) = 0

So, the values of x that satisfy the equation are x = 0 and x = 1.

4. To solve the equation (x - 2)√x - 3 = 0, we need to solve for x.

First, we need to eliminate the square root. We can do this by isolating it on one side.

(x - 2)√x = 3

Next, we'll square both sides of the equation to eliminate the square root.

[(x - 2)√x]² = (3)² (x - 2)²(x) = 9

Expand the left side:

(x² - 4x + 4)(x) = 9 x³ - 4x² + 4x = 9

Rearrange the equation:

x³ - 4x² + 4x - 9 = 0

At this point, we can either use numerical methods or factoring to find the exact solution. Unfortunately, factoring this cubic equation is not straightforward. Therefore, we can use numerical methods, such as graphing or using a calculator, to approximate the solution.