Question

x (x - 1) + x ( x + 3)=0

\(\dfrac{x}{2x-6}\)- \(\dfrac{x}{2x+2}\)=\(\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)

Answer 1

`x (x - 1) + x ( x + 3)=0`

`<=> x^2 - x + x^2 +3x=0`

`<=> 2x^2 +2x=0`

`<=> 2x(x+1)=0`

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{0;-1\right\}\)

__

\(\dfrac{x}{2x-6}-\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}-\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x-3\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-1\end{matrix}\right.\)

Ta có : \(\dfrac{x}{2\left(x-3\right)}-\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}-\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{2x.2}{2\left(x+1\right)\left(x-3\right)}\)

`=> x(x+1) -x(x-3)=4x`

`<=> x^2 + x -(x^2 -3x)=4x`

`<=> x^2 +x-x^2+3x-4x=0`

`<=>0=0`

 

Answer 2

\(x\left(x-1\right)+x\left(x+3\right)=0\)

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

\(\Leftrightarrow2x=0\)

\(\Leftrightarrow x=0\)

\(\dfrac{x}{2x-6}-\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\left(ĐKXĐ:x\ne-1;x\ne3\right)\)

\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}-\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}-\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{2.2x}{2\left(x+1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow\dfrac{x^2+x}{2\left(x+1\right)\left(x-3\right)}-\dfrac{x^2-3x}{2\left(x+1\right)\left(x-3\right)}-\dfrac{4x}{2\left(x+1\right)\left(x-3\right)}=0\)

\(\Rightarrow x^2+x-x^2+3x-4x=0\)

\(\Leftrightarrow0=0\)