Question

корінь квадратний х+2 - корінь квадратний 2х-3=1 ПОМОЖІТЬ ПОЖАЛУСТА

Answer 1

один в квадрате

1*2+2

1*2*2-3=1

 

 

 

 

Answer 2

I can help you solve this equation. First, let's rewrite it using LaTeX:

$$\sqrt{x+2}-\sqrt{2x-3}=1$$

To solve this equation, we need to isolate one of the square roots on one side and then square both sides. This will give us:

$$\sqrt{x+2}=1+\sqrt{2x-3}$$ $$(\sqrt{x+2})^2=(1+\sqrt{2x-3})^2$$

Expanding the squares and simplifying, we get:

$$x+2=1+2\sqrt{2x-3}+2x-3$$ $$-x+4=2\sqrt{2x-3}$$ $$(-x+4)^2=4(2x-3)$$

Expanding and simplifying again, we get:

$$x^2-8x+16=8x-12$$ $$x^2-16x+28=0$$

This is a quadratic equation that we can solve by using the quadratic formula:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

In this case, $a=1$, $b=-16$, and $c=28$. Plugging these values into the formula, we get:

$$x=\frac{16\pm\sqrt{(-16)^2-4(1)(28)}}{2(1)}$$ $$x=\frac{16\pm\sqrt{256-112}}{2}$$ $$x=\frac{16\pm\sqrt{144}}{2}$$ $$x=\frac{16\pm 12}{2}$$

This gives us two possible solutions:

$$x=\frac{16+12}{2}=14$$ $$x=\frac{16-12}{2}=2$$

However, we need to check if these solutions satisfy the original equation. If we plug in $x=14$, we get:

$$\sqrt{14+2}-\sqrt{2(14)-3}=1$$ $$\sqrt{16}-\sqrt{25}=1$$ $$4-5=1$$ $$-1=1$$

This is a false statement, so $x=14$ is not a valid solution. This is called an extraneous solution, which sometimes occurs when we square both sides of an equation.

If we plug in $x=2$, we get:

$$\sqrt{2+2}-\sqrt{2(2)-3}=1$$ $$\sqrt{4}-\sqrt{1}=1$$ $$2-1=1$$ $$1=1$$

This is a true statement, so $x=2$ is a valid solution. This is the only solution to the original equation.

Therefore, the final answer is:

$$x=2$$

I hope this helps you understand how to solve this equation. You can find more examples and explanations of solving equations with square roots on these websites.