Question

[Âu Nguyễn Bảo Châu]

Giải bài toán

Answer 1

[Nguyễn Huy Tú]

\(\frac{x}{x+1}-\frac{2x-3}{1-x}=\frac{3x^2+5}{x^2-1}\)ĐK : \(x\ne\pm1\)

\(\Leftrightarrow\frac{x}{x+1}+\frac{2x-3}{x-1}=\frac{3x^2+5}{x^2-1}\)

\(\Leftrightarrow\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\left(2x-3\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{3x^2+5}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow x^2-x+2x^2+2x-3x+3=3x^2+5\)

\(\Leftrightarrow-2x-2=0\Leftrightarrow x=-1\) vô lí 

Vậy phương trình vô nghiệm 

Answer 2

[Ngô Chi Lan]

\(\frac{x}{x+1}-\frac{2x-3}{1-x}=\frac{3x^2+5}{x^2-1}\)

\(\Leftrightarrow\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{\left(2x-3\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{3x^2+5}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow x^2-x+2x^2+2x-3x-3=3x^2+5\)

\(\Leftrightarrow x^2-x+2x^2+2x-3x-3-3x^2-5=0\)

\(\Leftrightarrow-2x-8=0\)

\(\Leftrightarrow-2x=8\)

\(\Leftrightarrow x=-4\)