Giải thích các bước giải:
d.ĐKXĐ: $x\ge -\dfrac12$
Ta có:
$\sqrt{2x+1}\ge x-1$
$\to 2\sqrt{2x+1}\ge 2x-2$
$\to 2\sqrt{2x+1}\ge (2x+1)-3$
$\to (2x+1)-2\sqrt{2x+1}-3\le 0$
$\to (\sqrt{2x+1}-3)(\sqrt{2x+1}+1)\le 0$
$\to \sqrt{2x+1}-3\le 0$ vì $\sqrt{2x+1}+1>0$
$\to \sqrt{2x+1}\le 3$
$\to 2x+1\le 9$
$\to x\le 4$
Kết hợp đkxđ$\to -\dfrac12\le x\le 4$
e.ĐKXĐ: $x\ge \dfrac32$
Ta có:
$\sqrt{2x-3}<2-x$
$\to x-2+\sqrt{2x-3}<0$
$\to 2x-4+2\sqrt{2x-3}<0$
$\to 2x-3+2\sqrt{2x-3}-1<0$
$\to (2x-3)+2\sqrt{2x-3}+1-2<0$
$\to (2x-3)+2\sqrt{2x-3}+1<2$
$\to (\sqrt{2x-3}+1)^2<2$
$\to -\sqrt{2}<\sqrt{2x-3}+1<\sqrt{2}$
$\to -1-\sqrt{2}<\sqrt{2x-3}<-1+\sqrt{2}$
$\to \sqrt{2x-3}<-1+\sqrt{2}$ vì $\sqrt{2x-3}\ge 0>-1-\sqrt{2}$
$\to 2x-3<(-1+\sqrt{2})^2$
$\to x<3-\sqrt{2}$
Kết hợp đkxđ $\to \dfrac32\le x<3-\sqrt{2}$
g.ĐKXĐ: $-2\le x\le 3$
Ta có:
$3\sqrt{-x^2+x+6}+2(2x-1)\le 0$
$\to 3\sqrt{(3-x)(2+x)}+2((x+2)+(x-3))\le 0$
$\to 2(x+2)+3\sqrt{2+x}\cdot \sqrt{3-x}-2(3-x)\le 0$
$\to (2\sqrt{2+x}-\sqrt{3-x})(\sqrt{2+x}+2\sqrt{3-x})\le 0$
$\to 2\sqrt{2+x}-\sqrt{3-x}\le 0$ vì $\sqrt{2+x}+2\sqrt{3-x}>0$
$\to 2\sqrt{2+x}\le \sqrt{3-x}$
$\to 4(2+x)\le 3-x$
$\to x\le -1$
Kết hợp đkxđ $\to -2\le x\le -1$
