\(\Leftrightarrow\sqrt[3]{3x+1}+\sqrt[3]{5-x}=\sqrt[3]{4x-3}+\sqrt[3]{9-2x}\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{3x+1}=a\\\sqrt[3]{5-x}=b\\\sqrt[3]{4x-3}=c\\\sqrt[3]{9-2x}=d\end{matrix}\right.\)
Ta được: \(\left\{{}\begin{matrix}a+b=c+d\\a^3+b^3=c^3+d^3\end{matrix}\right.\)
TH1:
Nếu \(a+b=c+d=0\Leftrightarrow\sqrt[3]{3x+1}+\sqrt[3]{5-x}=\sqrt[3]{4x-3}+\sqrt[3]{9-2x}=0\)
\(\Rightarrow\left\{{}\begin{matrix}3x+1=-\left(5-x\right)\\4x-3=-\left(9-2x\right)\end{matrix}\right.\) \(\Rightarrow x=-3\)
TH2: nếu \(a+b=c+d\ne0\)
\(a+b=c+d\Leftrightarrow\left(a+b\right)^3=\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=c^3+d^3+3cd\left(c+d\right)\)
\(\Leftrightarrow ab\left(a+b\right)=cd\left(c+d\right)\) (do \(a^3+b^3=c^3+d^3\))
\(\Leftrightarrow ab=cd\) (do \(a+b=c+d\ne0\))
\(\Leftrightarrow\sqrt[3]{\left(3x+1\right)\left(5-x\right)}=\sqrt[3]{\left(4x-3\right)\left(9-2x\right)}\)
\(\Leftrightarrow\left(3x+1\right)\left(5-x\right)=\left(4x-3\right)\left(9-2x\right)\)
\(\Leftrightarrow5x^2-28x+32=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{8}{5}\end{matrix}\right.\)
Vậy \(x=\left\{-3;4;\dfrac{8}{5}\right\}\)
Cái cuối này căn bậc 2 hay căn bậc 3 em? Căn bậc 2 thì hơi nghi ngờ về khả năng giải được của pt này.
