\(\Leftrightarrow\sqrt[3]{4x^2-9x-3}-\sqrt[3]{2x^2-3x-2}=\sqrt[3]{3x^2-2x+2}-\sqrt[3]{x^2+4x+3}\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{4x^2-9x-3}=a\\\sqrt[3]{2x^2-3x-2}=b\\\sqrt[3]{3x^2-2x+2}=c\\\sqrt[3]{x^2+4x+3}=d\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}a-b=c-d\\a^3-b^3=c^3-d^3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a-b=c-d\\\left(a-b\right)\left(a^2+ab+b^2\right)=\left(c-d\right)\left(c^2+cd+d^2\right)\end{matrix}\right.\)
TH1: \(a-b=c-d=0\) \(\Leftrightarrow2x^2-6x-1=0\Leftrightarrow...\)
TH2: \(a-b=c-d\ne0\) \(\Rightarrow a^2+ab+b^2=c^2+cd+d^2\)
\(\Leftrightarrow\left(a-b\right)^2+4ab=\left(c-d\right)^2+4cd\)
\(\Leftrightarrow ab=cd\)
\(\Leftrightarrow\left(4x^2-9x-3\right)\left(2x^2-3x-2\right)=\left(3x^2-2x+2\right)\left(x^2+4x+3\right)\)
\(\Leftrightarrow x\left(5x^3-40x^2+10x+25\right)=0\)
\(\Leftrightarrow5x\left(x-1\right)\left(x^2-7x-5\right)=0\)
\(\Leftrightarrow...\)
Nhận thấy x=0 là nghiệm của PT
Xét x khác 0
\(PT\Leftrightarrow\sqrt[3]{x^2+4x+3}-\sqrt[3]{3x^2-2x+2}=\sqrt[3]{2x^2-3x-2}-\sqrt[3]{4x^2-9x-3}\)\(\Leftrightarrow\frac{-2x^2+6x+1}{\left(\sqrt[3]{x^2+4x+3}\right)^2+\sqrt[3]{\left(x^2+4x+3\right)\left(3x^2-2x+2\right)}+\left(\sqrt[3]{3x^2-2x+2}\right)^2}\)\(=\frac{-2x^2+6x+1}{\left(\sqrt[3]{2x^2-3x-2}\right)^2+\sqrt[3]{\left(2x^2-3x-2\right)\left(4x^2-9x-3\right)}+\left(\sqrt[3]{4x^2-9x-3}\right)}\)
\(\Leftrightarrow\left(-2x^2+6x+1\right)\left(....\right)=0\)(tự viết cái trong ngoặc nhaa :33 dài quá)
\(\Leftrightarrow x=\frac{3\pm\sqrt{11}}{2}\)
Vậy ......
