Question

\(\dfrac{x-1}{x}\)-\(\dfrac{1}{x+1}\)=\(\dfrac{2x-1}{x^2+x}\)
Giải phương trình giúp mình với

Answer 1

ĐKXĐ: x<>0; x<>-1

PT =>(x-1)(x+1)-x=2x-1

=>x^2-1-x=2x-1

=>x^2-x-2x=0

=>x(x-3)=0

=>x=0(loại) hoặc x=3(nhận)

Answer 2

\(ĐK: x\ne 0; x\ne-1\)

Khi đó: 

\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{1.x}{\left(x+1\right).x}-\dfrac{2x-1}{x\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-1}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}-\dfrac{2x-1}{x\left(x+1\right)}=0\\ \Leftrightarrow x^2-1-x-2x+1=0\\ \Leftrightarrow x^2-3x=0\\ \Leftrightarrow x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

Vậy PT có nghiệm duy nhất \(S=\left\{3\right\}\)

Answer 3

\(ĐKXĐ:x\ne0;x\ne-1\)

\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x+1\right)}-\dfrac{1.x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\\ \Rightarrow x^2-x+x-1-x=2x-1\\ \Leftrightarrow x^2-x+x-x-2x=-1+1\\ \Leftrightarrow x^2-3x=0\\ \Leftrightarrow x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{3\right\}\)