a)
`(x-1)^2 -1+x^2=(1-x)(x+3)`
`<=> (1-x)^2 - (1-x)(1+x)-(1-x)(x+3)=0`
`<=> (1-x)(1-x-1-x-x-3)=0`
`<=> (1-x)(-3x-3)=0`
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\-3x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
vậy phương trình có tập nghiệm `S={1;-1}`
b)
\(\dfrac{2}{x^2-4}-\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x-4}{x\left(x+2\right)}=0\) \(\left(ĐKXĐ:x\ne2;x\ne-2;x\ne0\right)\)
\(\Leftrightarrow\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x-4}{x\left(x+2\right)}=0\)
suy ra `2x-(x-1)(x+2)+(x-4)(x-2)=0`
`<=> 2x-x^2 -2x + x + 2 + x^2 - 2x - 4x + 8 =0`
`<=> -x^2 + x^2 +2x -2x + x -2x - 4x +2 + 8 =0`
`<=> -5x + 10=0`
`<=> -5x = -10`
`<=> x=2(ktmđ)`
vậy phương trình vô nghiệm
`a)(x-1)^2-1+x^2=(1-x)(x+3)`
`<=>(x-1)^2-(x-1)(x+1)+(x-1)(x+3)=0`
`<=>(x-1)(x-1-x-1+x+3)=0`
`<=>(x-1)(x+1)=0`
`<=>x=+-1`
`b)2/[x^2-4]-[x-1]/[x(x-2)]+[x-4]/[x(x+2)]=0 ` `ĐK: x \ne +-2`
`=>2x-(x-1)(x+2)+(x-4)(x-2)=0`
`<=>2x-x^2-2x+x+2+x^2-2x-4x+8=0`
`<=>-5x=-10`
`<=>x=2` (t/m)
