`5, (x-1)^2-1+x^2=(1-x)(x+3)`
`<=> (x-1)^2+(x^2-1)-(1-x)(x+3)=0`
`<=> (x-1)^2+(x-1)(x+1)+(x-1)(x+3)=0`
`<=> (x-1)(x-1+x+1+x+3)=0`
`<=> (x-1)(3x+3)=0`
`<=> 3(x-1)(x+1)=0`
`<=>`\(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
Vậy `x∈{+-1}`
`8, x^4-4x^3+3x^2+4x-4=0`
`<=> (x^4-4x^3+4x^2)-(x^2-4x+4)=0`
`<=> x^2(x^2-4x+4)-(x^2-4x+4)=0`
`<=> (x^2-4x+4)(x^2-1)=0`
`<=> (x-2)^2. (x-1)(x+1)=0`
`<=>`\(\left[ \begin{array}{l}x=2\\x=-1\\x=1\end{array} \right.\)
Vậy `x∈{2;+-1}`
`9, (6x+1)/(x^2-7x+10)+5/(x+2)=3/(x-5)` ĐKXĐ: `x\ne5;-2`
`<=> (6x+1)/((x-2)(x-5))+5/(x+2)-3/(x-5)=0`
`<=> ((6x+1)(x+2)+5(x-2)(x-5)-3(x-2)(x+2))/((x-2)(x+2)(x+5))=0`
`<=> (6x^2+12x+x+2+5(x^2-7x+10)-3(x^2-4))/((x-2)(x+2)(x+5))=0`
`=> 6x^2+13x+2+5x^2-35x+50-3x^2+12=0`
`<=> 8x^2-22x+64=0`
`<=> 8(x^2-11/4x+8)=0`
`<=> x^2-2.x. 11/8+121/64+391/64=0`
`<=> (x-11/8)^2+391/64=0` (vô lý)
Vậy `x∈∅`
`10, 1/(3-x)-1/(x+1)=x/(x-3)-(x-1)^2/(x^2-2x-3)` ĐKXĐ: `x\ne3;-1`
`<=> 1/(3-x)-1/(x+1)-x/(x-3)+(x-1)^2/((x-3)(x+1))=0`
`<=> (-x-1-(x-3)-x(x+1)+(x-1)^2)/((x-3)(x+1))=0`
`=>-x-1-x+3-x^2-x+x^2-2x+1=0`
`<=> -5x+3=0`
`<=> x=3/5` (TM)
Vậy `x=3/5`
`11, 2/(x+2)-(2x^2+16)/(x^3+8)=5/(x^2-2x+4)` ĐKXĐ: `x\ne-2`
`<=> 2/(x+2)-(2x^2+16)/((x+2)(x^2-2x+4))-5/(x^2-2x+4)=0`
`<=> (2(x^2-2x+4)-(2x^2+16)-5(x+2))/((x+2)(x^2-2x+4)=0`
`=> 2x^2-4x+8-2x^2-16-5x-10=0`
`<=> -9x-18=0`
`<=> x=-2 (l)`
Vậy `x∈∅`
`13, 1/(x-2)-12/(x+3)=5/(6-x^2-x)`ĐKXĐ: `x\ne2;-3`
`<=> 1/(x-2)-12/(x+3)-5/(6-x^2-x)=0`
`<=>1/(x-2)-12/(x+3)+5/(x^2+x-6)=0`
`<=> (x+3-12(x-2)+5)/((x-2)(x+3))=0`
`=> x+3-12x+24+5=0`
`<=> -11x+32=0`
`<=> x=32/11` (TM)
Vậy `x=32/11`
`14 (x+1)/(x^2+x+1)-(x-1)/(x^2+x+1)=(2(x+2)^2)/(x^6-1)` ĐKXĐ: `x\ne+-1`)
`<=> (x+1-x+1)/(x^2+x+1)=(2(x+2)^2)/((x^3-1)(x^3+1))`
`<=> 2/(x^2+x+1)-(2(x+2)^2)/((x-1)(x^2+x+1)(x^3+1))=0`
`<=> (2(x-1)(x^3+1)-2(x^2+4x+4))/((x-1)(x^2+x+1)(x^3+1))=0`
`=> (2x-2)(x^3+1)-2x^2-8x-8=0`
`<=> 2x^4+2x-2x^3-2-2x^2-8x-8=0`
`<=> 2x^4-2x^3-2x^2-6x-10=0`
`<=> x^4-x^3-x^2-3x-5=0` (._?)
