\(\sqrt[3]{25+x}+\sqrt[3]{3-x}=4\)
\(pt\Leftrightarrow\sqrt[3]{25+x}-\left(\dfrac{x}{13}+\dfrac{37}{13}\right)+\sqrt[3]{3-x}-\left(-\dfrac{x}{13}+\dfrac{15}{13}\right)=0\)
\(\Leftrightarrow\dfrac{25+x-\left(\dfrac{x}{13}+\dfrac{37}{13}\right)^3}{\sqrt[3]{\left(25+x\right)^2}+\left(\dfrac{x}{13}+\dfrac{37}{13}\right)^2+\sqrt[3]{25+x}\left(\dfrac{x}{13}+\dfrac{37}{13}\right)}+\dfrac{3-x-\left(-\dfrac{x}{13}+\dfrac{15}{13}\right)^3}{\sqrt[3]{\left(3-x\right)^2}+\left(-\dfrac{x}{13}+\dfrac{15}{13}\right)^2+\sqrt[3]{3-x}\left(-\dfrac{x}{13}+\dfrac{15}{13}\right)}=0\)
\(\Leftrightarrow\dfrac{-\dfrac{\left(x-2\right)\left(x+24\right)\left(x+89\right)}{2197}}{\sqrt[3]{\left(25+x\right)^2}+\left(\dfrac{x}{13}+\dfrac{37}{13}\right)^2+\sqrt[3]{25+x}\left(\dfrac{x}{13}+\dfrac{37}{13}\right)}+\dfrac{\dfrac{\left(x-2\right)\left(x+24\right)\left(x-67\right)}{2197}}{\sqrt[3]{\left(3-x\right)^2}+\left(-\dfrac{x}{13}+\dfrac{15}{13}\right)^2+\sqrt[3]{3-x}\left(-\dfrac{x}{13}+\dfrac{15}{13}\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x-2\right)\left(x+24\right)}{2197}\left(\dfrac{-\left(x+89\right)}{\sqrt[3]{\left(25+x\right)^2}+\left(\dfrac{x}{13}+\dfrac{37}{13}\right)^2+\sqrt[3]{25+x}\left(\dfrac{x}{13}+\dfrac{37}{13}\right)}+\dfrac{x-67}{\sqrt[3]{\left(3-x\right)^2}+\left(-\dfrac{x}{13}+\dfrac{15}{13}\right)^2+\sqrt[3]{3-x}\left(-\dfrac{x}{13}+\dfrac{15}{13}\right)}\right)=0\)
\(\Rightarrow\dfrac{\left(x-2\right)\left(x+24\right)}{2197}=0\Rightarrow\)\(\left[{}\begin{matrix}x-2=0\\x+24=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-24\end{matrix}\right.\)
Cách kia khùng quá :|
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{25+x}=a\\\sqrt[3]{3-x}=b\end{matrix}\right.\)\(\Rightarrow a^3+b^3=28\)
Ta có hpt \(\left\{{}\begin{matrix}a+b=4\\a^3+b^3=28\end{matrix}\right.\)
ok thay hết vào nhau...
