`x^2(x-5)-4(x-5)=0`
`<=>(x-5)(x^2-4)=0`
`<=>(x-5)(x-2)(x+2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\\x=-2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{5;2;-2\right\}\)
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\(\dfrac{x+3}{x-3}-\dfrac{3}{x\left(x-3\right)}=\dfrac{1}{x}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-3\ne0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne0\end{matrix}\right.\)
Ta có : \(\dfrac{x+3}{x-3}-\dfrac{3}{x\left(x-3\right)}=\dfrac{1}{x}\)
\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x-3\right)}-\dfrac{3}{x\left(x-3\right)}=\dfrac{x-3}{x\left(x-3\right)}\)
`=> x^2 +3x -3 -x+3=0`
`<=> x^2 +2x=0`
`<=> x(x+2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-2\end{matrix}\right.\)
Vậy phương trình có nghiệm `x=-2`
\(a,x^2\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{5;\pm2\right\}\)
\(b,\dfrac{x+3}{x-3}-\dfrac{3}{x\left(x-3\right)}=\dfrac{1}{x}\left(ĐKXĐ:x\ne3;x\ne0\right)\)
\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x-3\right)}-\dfrac{3}{x\left(x-3\right)}-\dfrac{x-3}{x\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+3x-3-x+3}{x\left(x-3\right)}=0\)
\(\Rightarrow x^2+2x=0\)
\(\Leftrightarrow x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=-2\left(tm\right)\)
Vậy \(S=\left\{-2\right\}\)
