Giải:
a) \(\left(x-2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)
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b) \(\dfrac{-1}{1-x}-\dfrac{7}{x+2}=\dfrac{3}{\left(x-1\right)\left(x+2\right)}\) (1)
ĐKXĐ: \(x\ne1;x\ne-2\)
\(\left(1\right)\Leftrightarrow\dfrac{1}{x-1}-\dfrac{7}{x+2}=\dfrac{3}{\left(x-1\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{1\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}-\dfrac{7\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{3}{\left(x-1\right)\left(x+2\right)}\)
\(\Leftrightarrow1\left(x+2\right)-7\left(x-1\right)=3\)
\(\Leftrightarrow x+2-7x+7=3\)
\(\Leftrightarrow-6x+9=3\)
\(\Leftrightarrow-6x=-6\)
\(\Leftrightarrow x=1\) (Không thoả mãn ĐKXĐ)
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c) \(10x-2\left(2x+4\right)=2x\)
\(\Leftrightarrow10x-4x-8=2x\)
\(\Leftrightarrow10x-4x-2x=8\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
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a, \(\left(x-2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{2;1;-1\right\}\)
b, \(\dfrac{-1}{1-x}-\dfrac{7}{x+2}=\dfrac{3}{\left(x-1\right)\left(x+2\right)}\)(ĐKXĐ: \(x\ne1;x\ne-2\))
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{7}{x+2}=\dfrac{3}{\left(x-1\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{x+2-7\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{3}{\left(x-1\right)\left(x+2\right)}\)
\(\Rightarrow x+2-7\left(x-1\right)=3\)
\(\Leftrightarrow x+2-7x+7=3\)
\(\Leftrightarrow x-7x=3-2-7\)
\(\Leftrightarrow-6x=-6\)
\(\Leftrightarrow x=1\)(KTMĐK)
Vậy phương trình vô nghiệm.
c, \(10x-2\left(2x+4\right)=2x\)
\(\Leftrightarrow10x-4x-8=2x\)
\(\Leftrightarrow10x-4x-2x=8\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
Vậy phương trình có nghiệm là \(x=2\)
a,\(\left(x-2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
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b,ĐKXĐ:\(x\ne1;x\ne-2\)
\(-\dfrac{1}{1-x}-\dfrac{7}{x+2}=\dfrac{3}{\left(x-1\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{x+2}{x-1}-\dfrac{7\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{3}{\left(x-1\right)\left(x+2\right)}\)
\(\Leftrightarrow x+2-7x+7=3\)
\(\Leftrightarrow-6x+9=3\)
\(\Leftrightarrow-6x=-6\)
\(\Leftrightarrow x=1\) (Không tm)
Vậy phương trình vô nghiệm.
c,\(10x-2\left(2x+4\right)=2x\)
\(\Leftrightarrow10x-4x-8=2x\)
\(\Leftrightarrow10x-4x-2x=8\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
Vậy...
