ĐK: `x \ne 0; x \ne 1`
`x^2-x-18+72/(x^2-x)=0`
Đặt `t=x^2-x (t \ne 0;1)`
`t-18+72/t=0`
`<=>t^2-18t+72=0`
`<=> [(t=12),(t=6):}`
`<=> [(x^2-x=12),(x^2-x=6):}`
`<=> [( [(x=4),(x=-3):} ),( [(x=3),(x=-2):}):}`
Vậy `S={4;-3;3;-2}`.
b.
ĐKXĐ: \(x\ne\left\{-1;-3;-8;-10\right\}\)
\(\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{5}{\left(x+3\right)\left(x+8\right)}+\dfrac{2}{\left(x+8\right)\left(x+10\right)}=\dfrac{9}{52}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+10}=\dfrac{9}{52}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+10}=\dfrac{9}{52}\)
\(\Leftrightarrow\dfrac{9}{\left(x+1\right)\left(x+10\right)}=\dfrac{9}{52}\)
\(\Rightarrow\left(x+1\right)\left(x+10\right)=52\)
\(\Leftrightarrow x^2+11x-42=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-14\end{matrix}\right.\) (thỏa mãn)
a.
ĐKXĐ: \(x\ne\left\{0;1\right\}\)
Đặt \(x^2-x=t\) ta được:
\(t-18+\dfrac{72}{t}=0\)
\(\Rightarrow t\left(t-18\right)+72=0\)
\(\Leftrightarrow t^2-18t+72=0\)
\(\Leftrightarrow\left(t-6\right)\left(t-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=12\\t=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x=12\\x^2-x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-12=0\\x^2-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\\x=3\\x=-2\end{matrix}\right.\)
c.
ĐKXĐ: \(x\ne-2\)
\(x^2+\left(\dfrac{2x}{x+2}\right)^2-\dfrac{4x^2}{x+2}+\dfrac{4x^2}{x+2}=12\)
\(\Leftrightarrow\left(x-\dfrac{2x}{x+2}\right)^2+\dfrac{4x^2}{x+2}-12=0\)
\(\Leftrightarrow\left(\dfrac{x^2}{x+2}\right)^2+\dfrac{4x^2}{x+2}-12=0\)
Đặt \(\dfrac{x^2}{x+2}=t\)
\(\Rightarrow t^2+4t-12=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x^2}{x+2}=2\\\dfrac{x^2}{x+2}=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4x-4=0\\x^2+6x+12=0\left(vô-nghiệm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=2\pm2\sqrt{2}\)
