a) \(\dfrac{3}{x+1}-\dfrac{1}{x-2}=\dfrac{9}{\left(x+1\right)\left(2-x\right)}\)ĐKXĐ : \(x\ne-1;2\)
\(\Leftrightarrow\dfrac{3\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-2\right)}=\dfrac{-9}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow3x-6-x-1=-9\)
\(\Leftrightarrow2x=-2\)
\(\Leftrightarrow x=-1\)(không t/m ĐKXĐ )
Vậy....
b) \(\dfrac{x-4}{x-1}+\dfrac{x+4}{x+1}=2\)ĐKXĐ : \(x\ne\pm1\)
\(\Leftrightarrow\dfrac{\left(x-4\right)\left(x+1\right)}{x^2-1}+\dfrac{\left(x+4\right)\left(x-1\right)}{x^2-1}=\dfrac{2\left(x^2-1\right)}{x^2-1}\)
\(\Rightarrow x^2-3x-4+x^2+3x-4=2x^2-2\)
\(\Leftrightarrow\)\(6=0\)( vô lí )
Vậy pt vô nghiệm
a) ĐKXĐ: \(x\ne-1,x\ne2\)
Phương trình tương đương:
\(3\left(x-2\right)-\left(x+1\right)+9=0\\ \Leftrightarrow3x-6-x-1+9=0\\ \Leftrightarrow2x+2=0\\ \Leftrightarrow2x=-2\\ \Leftrightarrow x=-1\left(KTM\right)\)
Vậy \(S=\left\{\varnothing\right\}\)
b) ĐKXĐ: \(x\ne\pm1\)
Phương trình tương đương:
\(\left(x-4\right)\left(x+1\right)+\left(x+4\right)\left(x-1\right)=2\left(x-1\right)\left(x+1\right)\\ \Leftrightarrow2x^2-8=2\left(x^2-1\right)\\ \Leftrightarrow2x^2-8=2x^2-2\\ \Leftrightarrow2x^2-2x^2=-2+8\\ \Leftrightarrow0x=6\left(VN\right)\)
Vậy phương trình vô nghiệm
