giúp t vs mn $\frac{x-2\sqrt x+1}{\sqrt x-1}-\frac{x-\sqrt x}{\sqrt x-1}~~1)~\frac{x-25}{\sqrt x+5}-\frac{x+\sqrt x}{\sqrt x+1}$,$2)~\sqrt{(5-\sqrt3)^2}+\sqrt{(\sqrt3+1)^2}$ $3)~\sqrt{7-2\sqrt{10}}-\sqrt{(\sqrt2+\sqrt5)^2}$

Question

Accepted Answer

1) $\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}-\frac{x-\sqrt{x}}{\sqrt{x}-1}$

Điều kiện xác định: $x \geq 0$ và $x \neq 1$.

Ta có:
$$
\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}-\frac{x-\sqrt{x}}{\sqrt{x}-1} = \frac{(\sqrt{x}-1)^2-(x-\sqrt{x})}{\sqrt{x}-1}
$$
$$
= \frac{x - 2\sqrt{x} + 1 - x + \sqrt{x}}{\sqrt{x}-1} = \frac{-\sqrt{x} + 1}{\sqrt{x}-1} = \frac{-(\sqrt{x} - 1)}{\sqrt{x}-1} = -1
$$

2) $\frac{x-25}{\sqrt{x}+5}-\frac{x+\sqrt{x}}{\sqrt{x}+1}$

Điều kiện xác định: $x \geq 0$ và $x \neq 25$.

Ta có:
$$
\frac{x-25}{\sqrt{x}+5}-\frac{x+\sqrt{x}}{\sqrt{x}+1} = \frac{(\sqrt{x}+5)(\sqrt{x}-5)}{\sqrt{x}+5} - \frac{x+\sqrt{x}}{\sqrt{x}+1}
$$
$$
= \sqrt{x} - 5 - \frac{x+\sqrt{x}}{\sqrt{x}+1}
$$

3) $\sqrt{(5-\sqrt{3})^2}+\sqrt{(\sqrt{3}+1)^2}$

Ta có:
$$
\sqrt{(5-\sqrt{3})^2} = |5-\sqrt{3}| = 5 - \sqrt{3} \quad (\text{vì } 5 > \sqrt{3})
$$
$$
\sqrt{(\sqrt{3}+1)^2} = |\sqrt{3}+1| = \sqrt{3} + 1
$$
$$
\sqrt{(5-\sqrt{3})^2} + \sqrt{(\sqrt{3}+1)^2} = (5 - \sqrt{3}) + (\sqrt{3} + 1) = 6
$$

4) $\sqrt{7-2\sqrt{10}}-\sqrt{(\sqrt{2}+\sqrt{5})^2}$

Ta có:
$$
\sqrt{7-2\sqrt{10}} = \sqrt{(\sqrt{5}-\sqrt{2})^2} = |\sqrt{5}-\sqrt{2}| = \sqrt{5} - \sqrt{2} \quad (\text{vì } \sqrt{5} > \sqrt{2})
$$
$$
\sqrt{(\sqrt{2}+\sqrt{5})^2} = |\sqrt{2}+\sqrt{5}| = \sqrt{2} + \sqrt{5}
$$
$$
\sqrt{7-2\sqrt{10}} - \sqrt{(\sqrt{2}+\sqrt{5})^2} = (\sqrt{5} - \sqrt{2}) - (\sqrt{2} + \sqrt{5}) = -2\sqrt{2}
$$

Đáp số:
1) $-1$
2) $\sqrt{x} - 5 - \frac{x+\sqrt{x}}{\sqrt{x}+1}$
3) $6$
4) $-2\sqrt{2}$