\(d.\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)
Gọi \(t=x^2-5x\), ta được:
\(t^2+10t+24=0\)
\(\Leftrightarrow\left(t+4\right)\left(t+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-4\\t=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x=-4\\x^2-5x=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+4=0\\x^2-5x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\\\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\end{matrix}\right.\)
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\(e.\dfrac{1}{x+2}+\dfrac{1}{x^2-2x}=\dfrac{8}{x^3-4x}\left(x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{x\left(x-2\right)}{x\left(x^2-4\right)}+\dfrac{x+2}{x\left(x^2-4\right)}=\dfrac{8}{x\left(x^2-4x\right)}\)
\(\Leftrightarrow x^2-2x+x+2=8\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow x^2+2x-3x-6=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
d) Đặt t = \(x^2-5x\) ; p/t : \(t^2+10t+24=0\) \(\Leftrightarrow\left(t+4\right)\left(t+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-4\\t=-6\end{matrix}\right.\) hay \(\left[{}\begin{matrix}x^2-5x=-4\left(1\right)\\x^2-5x=-6\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^2-5x+4=0\) \(\Leftrightarrow\left(x-4\right)\left(x-1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow x^2-5x+6=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
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