Phương trình bậc nhất một ẩn

H24

Giải phương trình sau:

\(\left(\dfrac{x+3}{x-2}\right)^2+6\left(\dfrac{x-3}{x+2}\right)^2=\dfrac{7\left(x^2-9\right)}{x^2-4}\)

NL
16 tháng 2 2019 lúc 20:39

\(x\ne\pm2\)

\(\left(\dfrac{x+3}{x-2}\right)^2+6\left(\dfrac{x-3}{x+2}\right)^2-7.\dfrac{\left(x-3\right)}{\left(x+2\right)}.\dfrac{\left(x+3\right)}{\left(x-2\right)}=0\)

\(\Leftrightarrow\left(\dfrac{x+3}{x-2}\right)^2-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x+2\right)}-6\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x+2\right)\left(x-2\right)}+6\left(\dfrac{x-3}{x+2}\right)^2=0\)

\(\Leftrightarrow\dfrac{x+3}{x-2}\left(\dfrac{x+3}{x-2}-\dfrac{x-3}{x+2}\right)-6\dfrac{x-3}{x+2}\left(\dfrac{x+3}{x-2}-\dfrac{x-3}{x+2}\right)=0\)

\(\Leftrightarrow\left(\dfrac{x+3}{x-2}-\dfrac{6\left(x-3\right)}{x+2}\right)\left(\dfrac{x+3}{x-2}-\dfrac{x-3}{x+2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x+3}{x-2}=\dfrac{6\left(x-3\right)}{x+2}\\\dfrac{x+3}{x-2}=\dfrac{x-3}{x+2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+6=6\left(x^2-5x+6\right)\\x^2+5x+6=x^2-5x+6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x^2-35x+30=0\\5x=-5x\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=6\\x=0\end{matrix}\right.\)

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NH
16 tháng 2 2019 lúc 20:41

\(\left(\dfrac{x+3}{x-2}\right)^2+6\left(\dfrac{x-3}{x+2}\right)^2=\dfrac{7\left(x^2-9\right)}{x^2-4}\)

\(\Leftrightarrow\dfrac{\left(x+3\right)^2}{\left(x-2\right)^2}\left(x-2\right)^2\left(x+2\right)+\dfrac{6\left(x-3\right)^2}{\left(x+2\right)^2}\left(x-2\right)^2\left(x+2\right)^2=\dfrac{7\left(x^2-9\right)}{x^2-4}\left(x-2\right)^2\left(x+2\right)^2\)

\(\Leftrightarrow\left(x+3\right)^2\left(x+2\right)^2+6\left(x-3\right)^2\left(x-2\right)^2=7\left(x-2\right)\left(x+2\right)\left(x+3\right)\left(x-3\right)\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=6\end{matrix}\right.\)

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