`b)1/x^2+1/(4-x)^2=1/2` `(đk: x\ne0;x\ne4)`

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

`2(16-8x+x^2)+2x^2=16x^2-8x^3+x^4`

`32-16x+4x^2=16x^2-8x^3+x^4`

`x^4-8x^3+12x^2+16x-32=0`

`(x^2-4x)^2-4(x^2-4x)-32=0`

`(x^2-4x-8)(x^2-4x+4)=0`

`TH1:`

`x^2-4x-8=0`

`\Delta=(-4)^2-4.1.(-8)=48`

`x=(4+-\sqrt48)/2`

`x=(4+-4\sqrt3)/2`

`x=2+-2\sqrt3`

`TH2:`

`x^2-4x+4=0`

`(x-2)^2=0`

`x=2`

Vậy `S={2-2\sqrt3;2;2+2\sqrt3}`

`c)1/x^2+1/(x+1)^2=5/4` `(đk: x\ne0;x\ne-1)`

`4(x+1)^2+4x^2=5x^2(x+1)^2`

`4(x^2+2x+1)+4x^2=5x^2(x^2+2x+1)`

`8x^2+8x+4=5x^4+10x^3+5x^2`

`5x^4+10x^3-3x^2-8x-4=0`

`5(x^2+x)^2-8(x^2+x)-4=0`

`(5x^2+5x+2)(x^2+x-2)=0`

`TH1:`

`5x^2+5x+2=0`

`\Delta=5^2-4.5.2=-15<0` `(`vô nghiệm`)`

`TH2:`

`x^2+x-2=0`

`(x-1)(x+2)=0`

`x=1` hoặc `x=-2`

Vậy `S={-2;1}`