Đáp án:
Giải thích các bước giải:
1.
$⇔\sqrt[3]{3x+2}-1+3x^3+x^2+3x+1=0$
$⇔\dfrac{3x+1}{\sqrt[3]{(3x+2)^2}+\sqrt[3]{3x+2}+1}+x^2(3x+1)+3x+1=0$
$⇔(3x+1)(\dfrac{1}{\sqrt[3]{(3x+2)^2}+\sqrt[3]{3x+2}+1}+x^2+1)=0$
$⇔3x+1=0$ (ngoặc phía sau luôn dương)
$⇔x=-\dfrac{1}{3}$
Câu 2:
ĐKXĐ: $x \geq -\dfrac{13}{2}$
$⇔\sqrt[3]{2x+1}+2+\sqrt{2x+13}-2+(2x+9)(2x+13)=0$
$⇔\dfrac{2x+9}{\sqrt[3]{(2x+1)^2}-2\sqrt[3]{2x+1}+4}+\dfrac{2x+9}{\sqrt{2x+13}+2}+(2x+9)(2x+13)=0$
$⇔(2x+9)\left ( \dfrac{1}{\sqrt[3]{(2x+1)^2}-2\sqrt[3]{2x+1}+4}+\dfrac{1}{\sqrt{2x+13}+2}+2x+13 \right )=0$
$⇔2x+9=0$ (ngoặc phía sau luôn dương)
$⇔x=-\dfrac{9}{2}$
