\(\frac{3x-3}{x^2-1}=\frac{x}{x-2}-1\)ĐKXĐ : \(x\ne\pm1;x\ne2\)
\(\Leftrightarrow\frac{3\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x+1\right)\left(x-2\right)}=\frac{x\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}-\frac{\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\frac{3\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{x\left(x+1\right)-\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(\Rightarrow3x-6=x^2+x-x^2+x+2\)
\(\Leftrightarrow3x-6-2x-2=0\)
\(\Leftrightarrow x-8=0\)
\(\Leftrightarrow x=8\)( thỏa )
Vậy....
\(\frac{3x-3}{x^2-1}=\frac{x}{x-2}-\)\(1\)
\(\Leftrightarrow\) \(\frac{3.\left(x-1\right)}{\left(x-1\right).\left(x+1\right)}\)\(=\frac{x}{x-2}-1\)
\(\Leftrightarrow\)\(\frac{3}{x+1}=\frac{x}{x-2}-1\)
ĐKXĐ : \(x\ne-1,2\)
\(\Leftrightarrow\)\(\frac{3.\left(x-2\right)}{\left(x+1\right).\left(x-2\right)}\)\(=\frac{x.\left(x+1\right)}{\left(x+1\right).\left(x-2\right)}\)\(-\frac{\left(x+1\right).\left(x-2\right)}{\left(x+1\right).\left(x-2\right)}\)
\(\Leftrightarrow\)\(3x-6=x^2+x-\left(x^2-2x+x-2\right)\)
\(\Leftrightarrow\)\(3x-6=x^2+x-x^2+x+2\)
\(\Leftrightarrow\)\(3x-x-x=6+2\)
\(\Leftrightarrow\) \(x=8\)
Vậy phương trình có nghiệm là : \(x=8\)
