`#0409`

`4,`

`(x-3)/(x-2)+(x-2)/(x-4)=-1`    ĐK `:x\ne2;x\ne4`

`<=>((x-3)(x-4))/((x-2)(x-4))+(x-2)^2/((x-2)(x-4))=-1`

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

`<=>x^2-7x+12+x^2-4x+4=-(x^2-6x+8)`

`<=>2x^2-11x+16=-x^2+6x-8`

`<=>2x^2-11x+16+x^2-6x+8=0`

`<=>3x^2-17x+24=0`

`<=>3x^2-9x-8x+24=0`

`<=>3x(x-3)-8(x-3)=0`

`<=>(3x-8)(x-3)=0`

`th_1 : 3x-8=0`

`<=>3x=8`

`<=>x=8/3` `(tm)`

`th_2 : x-3=0`

`<=>x=3` `(tm)`

Vậy `S={8/3;3}`

`5,`

`(x+1)/(x-1)-(x-1)/(x+1)=4/((x-1)(x+1))`     ĐK `:x\ne+-1`

`<=>(x+1)^2/((x-1)(x+1))-(x-1)^2/((x-1)(x+1))=4/((x-1)(x+1))`

`=>(x+1)^2-(x-1)^2=4`

`<=>x^2+2x+1-(x^2-2x+1)=4`

`<=>x^2+2x+1-x^2+2x-1=4`

`<=>4x=4`

`<=>x=1` `(ktm)`

Vậy `S=`$\emptyset$

`6,`

`(x-3)/(x-2)-(x-2)/(x-4)=16/5`    ĐK`:x\ne2;x\ne4`

`<=>((x-3)(x-4))/((x-2)(x-4))-(x-2)^2/((x-2)(x-4))=16/5`

`<=>((x-3)(x-4)-(x-2)^2)/((x-2)(x-4))=16/5`

`<=>(x^2-7x+12-x^2+4x-4)/(x^2-6x+8)=16/5`

`<=>(-3x+8)/(x^2-6x+8)=16/5`

`<=>5(-3x+8)=16(x^2-6x+8)`

`<=>-15x+40-16x^2+96x-128=0`

`<=>-16x^2+81x-88=0` `(**)`

Xét `\Delta=81^2-4 . (-16) . (-88)=929`

Vì `\Delta>0` nên phương trình `(**)` có `2` nghiệm phân biệt `:`

`x_1=(-81+\sqrt{929})/(2 . (-16))=(81-\sqrt{929})/32` `(tm)`

`x_2=(-81-\sqrt{929})/(2 . (-16))=(81+\sqrt{929})/32` `(tm)`

Vậy `S={(81-\sqrt{929})/32;(81+\sqrt{929})/32}`

`7,`

`(x-2)/(x+2)-3/(x-2)=(2(x-11))/((x-2)(x+2))`   ĐK `:x\ne+-2`

`<=>(x-2)^2/((x-2)(x+2))-(3(x+2))/((x-2)(x+2))=(2(x-11))/((x-2)(x+2))`

`=>(x-2)^2-3(x+2)=2(x-11)`

`<=>x^2-4x+4-3x-6=2x-22`

`<=>x^2-7x-2-2x+22=0`

`<=>x^2-9x+20=0`

`<=>x^2-4x-5x+20=0`

`<=>x(x-4)-5(x-4)=0`

`<=>(x-5)(x-4)=0`

`th_1 : x-5=0`

`<=>x=5` `(tm)`

`th_2 : x-4=0`

`<=>x=4` `(tm)`

Vậy `S={5;4}`

`8,`

`(x-1)/(x+2)-x/(x-2)=(2-5x)/((x-2)(x+2))`   ĐK `:x\ne+-2`

`<=>((x-1)(x-2))/((x-2)(x+2))-(x(x+2))/((x-2)(x+2))=(2-5x)/((x-2)(x+2))`

`=>(x-1)(x-2)-x(x+2)=2-5x`

`<=>x^2-3x+2-x^2-2x-2+5x=0`

`<=>0=0` `(` luôn đúng `)`

Vậy phương trình có vô số nghiệm trừ `x=+-2`