Đáp án:
` S={1/2;-1/2}`
Giải thích các bước giải:
` (x/(x-1))^2+(x/(x+1))^2=10/9(x\ne +-1)`
`<=> (x/(x-1))^2+(x/(x+1))^2+2.(x/(x-1)).(x/(x+1))=10/9+(2x^2)/((x-1)(x+1))`
`<=>(x/(x-1)+x/(x+1))^2=10/9+(2x^2)/(x^2-1)`
`<=>((x(x+1)+x(x-1))/((x-1)(x+1)))^2=10/9+(2x^2)/(x^2-1)`
`<=>((x^2+x+x^2-x)/(x^2-1))^2=10/9+(2x^2)/(x^2-1)`
`<=>((2x^2)/(x^2-1))^2=10/9+(2x^2)/(x^2-1)`
Đặt `(2x^2)/(x^2-1)=t`, ta có:
`t^2=10/9+t`
`<=>t^2-t-10/9=0`
`<=>9t^2-9t-10=0`
`<=>9t^2-15t+6t-10=0`
`<=>3t(3t-5)+2(3t-5)=0`
`<=>(3t-5)(3t+2)=0<=>`\(\left[ \begin{array}{l}t=\dfrac{5}{3}\\t=\dfrac{-2}{3}\end{array} \right.\)
Nếu `t=5/3=> (2x^2)/(x^2-1)=5/3`
`<=>6x^2=5x^2-5`
`<=>x^2=-5=>x=∅(vì x^2>=0)`
Nếu`t=-2/3=> (2x^2)/(x^2-1)=-2/3`
`<=>6x^2=-2x^2+2`
`<=>8x^2=2`
`<=>x^2=1/4<=>x=+-1/2`(t/m)
Vậy` S={1/2;-1/2}`
