giải bất phương trình sau: 1) $\sqrt[]{2x+3}$ +$\sqrt[]{x+2}$ $\leq$ 1 2) $\sqrt[]{11-x}$ – $\sqrt[]{x-1}$ > 2

Giải thích các bước giải:

1)đkxđ $x\ge -\dfrac 32$

$\sqrt{2x+3}+\sqrt{x+2}\le 1$ 

$\to\sqrt{2x+3}\le 1-\sqrt{x+2}$ 

$\to 2x+3\le (1-\sqrt{x+2})^2$ 

$\to 2x+3\le 1+x+2-2\sqrt{x+2}$

$\to x\le -2\sqrt{x+2}$

$\to x^2\le 4(x+2)$

$\to -2\le x\le -2\sqrt{3}+2$

$\to -\dfrac 32 \le x\le -2\sqrt{3}+2$

2)Đkxđ $1\le x\le 11$

$\sqrt{11-x}-\sqrt{x-1}>2$

$\to \sqrt{11-x}>2+\sqrt{x-1}$

$\to 11-x>(2+\sqrt{x-1})^2$

$\to 11-x>4+4\sqrt{x-1}+x-1$

$\to 8-2x>4\sqrt{x-1}$

$\to 4-x>2\sqrt{x-1}$

$\to x-1+2\sqrt{x-1}-3<0$

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

$\to \sqrt{x-1}<1$

$\to x<2$

$\to 1\le x<2$