`a)10x-5=x(2x-1)`
`⇔5(2x-1)-x(2x-1)=0`
`⇔(2x-1)(5-x)=0`
⇔\(\left[ \begin{array}{l}2x=1\\x=5\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{1}{2}\\x=5\end{array} \right.\)
`b)x-5/x+x-3/x+5=x-25/x^2+5x` ĐK:x$\neq$ ,±5
`⇔(x-5)(x+5)/x(x+5)+x(x+3)/x(x+5)=x-25/x(x+5)`
`⇔x^2-5x+5x-25+x^2+3x-x=-25`
`⇔2x^2-4x=25-25`
`⇔2x(x-2)=0
`2x=0 hoặc x=2`
`x=0(KTM)`
`x=2(TM)`
Vậy phương trình vô nghiệm S={2}
`c)x-1/x+3-2x/x-3=7-6x/x^2-9` ĐK:x$\neq$ ±3
`⇔(x-1)(x-3)/x^2-9-2x(x+3)/x^2-9=7-6x/x^2-9`
`⇔x^2-x-3x+3-2x^2-6x=7-6x`
`⇔x^2-4x+3-7=0`
`⇔x^2-4x-4=0`
`⇔-(x+2)^2=0`
`⇔x=-2(TM)`
Xin hay nhất ạ.