Giải thích các bước giải:
d.ĐKXĐ: $x\ge -1$
Ta có:
$2x^2-x+3=3\sqrt{x^3+1}$
$\to 2(x^2-x+1)+(x+1)=3\sqrt{(x+1)(x^2-x+1)}$
$\to 2(x^2-x+1)-3\sqrt{(x+1)(x^2-x+1)}+(x+1)=0$
$\to (\sqrt{x^2-x+1}-\sqrt{x+1})(2\sqrt{x^2-x+1}-\sqrt{x+1})=0$
$\to \sqrt{x^2-x+1}-\sqrt{x+1}=0\to \sqrt{x^2-x+1}=\sqrt{x+1}$
$\to x^2-x+1=x+1\to x^2-2x=0\to x(x-2)=0\to x\in\{0,2\}$
Hoặc $2\sqrt{x^2-x+1}-\sqrt{x+1}=0\to 2\sqrt{x^2-x+1}=\sqrt{x+1}$
$\to 4(x^2-x+1)=x+1$
$\to$Vô nghiệm
e.Ta có:
$8x^3-4x-2=\sqrt[3]{6x+2}$
$\to (2x)^3+2x=(6x+2)+\sqrt[3]{6x+2}$
Đặt $2x=a, \sqrt[3]{6x+2}=b$
$\to a^3+a=b^3+b$
$\to (a^3-b^3)+(a-b)=0$
$\to (a-b)(a^2+ab+b^2)+(a-b)=0$
$\to (a-b)(a^2+ab+b^2+1)=0$
$\to a-b=0$ vì $a^2+ab+b^2+1>0$
$\to a=b$
$\to 2x=\sqrt[3]{6x+2}$
$\to 8x^3=6x+2$
$\to x\in\{1,-\dfrac12\}$
