Đáp án:
`1)` `A=-110`
`2)` `(x^2+y^2)(x^2-2y^2+1)`
Giải thích các bước giải:
`1)` Ta có:
`\qquad (x+1/x )^2=x^2+2.x. 1/x+1/{x^2}`
`=(x^2+1/{x^2})+2=23+2=25`
`=>(x+1/x)^2=25`
`=>`$\left[\begin{array}{l}x+\dfrac{1}{x}=5\\x+\dfrac{1}{x}=-5\end{array}\right.$
Vì `x<0=>1/x<0=>x+1/x<0`
`=>x+1/x=-5`
Ta có:
`\qquad (x+1/x)^3=(-5)^3`
`<=>x^3+3x^2 . 1/x +3x. 1/{x^2} +1/{x^3}=-125`
`<=>x^3+1/{x^3}+3(x+1/x)=-125`
`<=>x^3+1/{x^3}+3. (-5)=-125`
`<=>x^3+1/{x^3}=-125+15=-110`
Vậy `A=x^3+1/{x^3}=-110`
$\\$
`2)` `x^4-2y^4-x^2y^2+x^2+y^2`
`=(x^4-y^4)-(x^2y^2+y^4)+x^2+y^2`
`=(x^2-y^2)(x^2+y^2)-y^2(x^2+y^2)+x^2+y^2`
`=(x^2+y^2)(x^2-y^2-y^2+1)`
`=(x^2+y^2)(x^2-2y^2+1)`
Vậy: `x^4-2y^4-x^2y^2+x^2+y^2`
`=(x^2+y^2)(x^2-2y^2+1)`
