a.
Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:
\(\dfrac{4}{4x+\dfrac{7}{x}-8}+\dfrac{3}{4x+\dfrac{7}{x}-10}=1\)
Đặt \(4x+\dfrac{7}{x}-10=t\)
\(\Rightarrow\dfrac{4}{t+2}+\dfrac{3}{t}=1\)
\(\Leftrightarrow7t+6=t^2+2t\)
\(\Leftrightarrow t^2-5t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\dfrac{7}{x}-10=-1\\4x+\dfrac{7}{x}-10=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x^2-9x+7=0\\4x^2-16x+7=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
c.
ĐKXĐ: \(x\ne\pm1\)
\(\Leftrightarrow20\left(\dfrac{x-2}{x+1}\right)^2-5\left(\dfrac{x+2}{x-1}\right)+48\left(\dfrac{x-2}{x+1}\right)\left(\dfrac{x+2}{x-1}\right)=0\)
Đặt \(\left\{{}\begin{matrix}\dfrac{x-2}{x+1}=a\\\dfrac{x+2}{x-1}=b\end{matrix}\right.\)
\(\Rightarrow20a^2-5b^2+48ab=0\)
\(\Leftrightarrow\left(10a-b\right)\left(2a+5b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}10a=b\\2a=-5b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{10\left(x-2\right)}{x+1}=\dfrac{x+2}{x-1}\\\dfrac{2\left(x-2\right)}{x+1}=\dfrac{-5\left(x+2\right)}{x-1}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}10\left(x-2\right)\left(x-1\right)-\left(x+2\right)\left(x+1\right)=0\\2\left(x-2\right)\left(x-1\right)+5\left(x+2\right)\left(x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(x\ne-3\)
\(x^2+\left(\dfrac{3x}{x+3}\right)^2-\dfrac{6x^2}{x+3}+\dfrac{6x^2}{x+3}=40\)
\(\Leftrightarrow\left(x-\dfrac{3x}{x+3}\right)^2+\dfrac{6x^2}{x+3}-40=0\)
\(\Leftrightarrow\left(\dfrac{x^2}{x+3}\right)^2+\dfrac{6x^2}{x+3}-40=0\)
Đặt \(\dfrac{x^2}{x+3}=t\)
\(\Rightarrow t^2+6t-40=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x^2}{x+3}=4\\\dfrac{x^2}{x+3}=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4x-12=0\\x^2+10x+30=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
