Question

9 sqrt{x - 1}=x^2 sqrt{x - 1}

Answer 1

$9 \sqrt{x - 1}=x^2 \sqrt{x - 1}$

D:

$x \ge 1$

$x^2 \sqrt{x - 1}-9 \sqrt{x - 1}=0$

$\sqrt{x - 1}(x^2-9)=0$

$\sqrt{x - 1}(x-3)(x+3)=0$

$\sqrt{x - 1}=0\ \ \ \ \ x-3=0\ \ \ \ \ x+3=0$

$x=1\ \ \ \ \ x=3\ \ \ \ \ x=-3$

Ответ

x=1,x=3

Answer 2

To solve the equation 9√(x - 1) = x^2√(x - 1), you can start by dividing both sides of the equation by √(x - 1) to isolate the radical term:

9 = x^2

Now, you have a quadratic equation. To solve for x, you can take the square root of both sides:

√9 = √(x^2)

3 = x

So, the solution to the equation is x = 3. However, you should check this solution to ensure it is valid. When dealing with square roots, you need to make sure that the expression inside the square root is non-negative, or else it would result in imaginary solutions.

In this case, x - 1 must be non-negative:

x - 1 ≥ 0

x ≥ 1

Since x = 3 satisfies this condition, the solution x = 3 is valid.