\(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{7x-3}{9-x^2}\)ĐK : \(x\ne\pm3\)
\(\Leftrightarrow\frac{x-1}{x+3}+\frac{x}{3-x}=\frac{7x-3}{9-x^2}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(3-x\right)+x\left(x+3\right)}{\left(x+3\right)\left(3-x\right)}=\frac{7x-3}{\left(3-x\right)\left(x+3\right)}\)
\(\Rightarrow3x-x^2-3+x+x^2+3x=7x-3\)
\(\Leftrightarrow7x-3=7x-3\Leftrightarrow0x=0\)
Vậy phương trình có vô số nghiệm
Trả lời:
\(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{7x-3}{9-x^2}\)\(\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{3-7x}{x^2-9}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{x^2-9}-\frac{x\left(x+3\right)}{x^2-9}=\frac{3-7x}{x^2-9}\)
\(\Rightarrow x^2-3x-x+3-\left(x^2+3x\right)=3-7x\)
\(\Leftrightarrow x^2-4x+3-x^2-3x=3-7x\)
\(\Leftrightarrow3-7x=3-7x\)
\(\Leftrightarrow-7x+7x=3-3\)
\(\Leftrightarrow0x=0\)( luôn thỏa mãn )
Vậy \(S=ℝ\)với \(x\ne\pm3\)
