\(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)
\(\Leftrightarrow2x^2-6x=0\)
=>2x(x-3)=0
=>x=3(loại) hoặc x=0(nhận)
\(\dfrac{x}{2x-6}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)
\(ĐK:x\ne3;x\ne-1\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)+x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{4x}{2\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)
\(\Leftrightarrow x^2+x+x^2-3x-4x=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
`x/[2x-6]+x/[2x+2]=[2x]/[(x+1)(x-3)]` `ĐK: x \ne -1,x \ne 3`
`<=>x/[2(x-3)]+x/[2(x+1)]=[2x]/[(x+1)(x-3)]`
`<=>[x(x+1)+x(x-3)]/[2(x-3)(x+1)]=[4x]/[2(x+1)(x-3)]`
`=>x^2+x+x^2-3x=4x`
`<=>2x^2-6x=0`
`<=>2x(x-3)=0`
`<=>` $\left[\begin{matrix} x=0\text{ (t/m)}\\ x=3\text{ (ko t/m)}\end{matrix}\right.$
Vậy `S={0}`
đkxđ: \(x\ne-1;3\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\\ \Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{4x}{2\left(x+1\right)\left(x-3\right)}=0\\ \Rightarrow x\left(x+1\right)+x\left(x-3\right)-4x=0\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\)
vậy x = 0
