Chủ đề này có 1 trả lời

[thuy32]

(2x + 4)$\sqrt[3]{2x+3}$ - $\sqrt{9x³+60x²+133x+98}$=x²-2x-5

[nguyenquangtruonghktcute]

(2x + 4)$\sqrt[3]{2x+3}$ - $\sqrt{9x³+60x²+133x+98}$=x²-2x-5

$9x^3+60x^2+133x+98=(3x+7)^2(x+2)$ từ đó ta có ĐK $x\geq -2$

pt $\Leftrightarrow (\sqrt[3]{2x+3})^4+3(\sqrt[3]{2x+3})^3+\sqrt[3]{2x+3}=(\sqrt{x+2})^4+3(\sqrt{x+2})^3+\sqrt{x+2} \Leftrightarrow \sqrt[3]{2x+3}=\sqrt{x+2} \Leftrightarrow (x+1)(x^2+x-1)=0 \Leftrightarrow x=-1; x=\frac{-1+\sqrt{5}}{2}; x=\frac{-1-\sqrt{5}}{2}$ (loại)

$S=\left \{ -1;\frac{-1+\sqrt{5}}{2} \right \}$