`GPT` `1/(x^2+2x+2)^2+1/(x^2+2x+3)^2=5/4` 27/07/2021 Bởi Peyton `GPT` `1/(x^2+2x+2)^2+1/(x^2+2x+3)^2=5/4`
đặt `x^2+2x+3=a` `⇔1/(a-1)^2+1/a^2=5/4` `⇔(2a^2-2a+1)/((a^2-2a+1)a^2)=5/4` `⇔8a^2-8a+4=5a^4-10a^3+5a^2` `⇔-5x^4-10a^3+3a^2+8a+4=0` `⇔(1-a)(a+2)(5a^2+5a+2)=0` `⇔`\(\left[ \begin{array}{l}a=1\\a=-2\end{array} \right.\) `(vì 5a^2+5a+2>0)` `⇒x=-1` Bình luận
Đặt `x^2+2x+2=t` PT `<=>1/t^2 + 1/(t+1)^2=5/4` `<=>1/t^2 + 1/(t^2+2t+1)=5/4` `<=>(2t^2+2t+1)/(t^2(t^2+2t+1))=5/4` `<=>8t^2+8t+4=5t^4+10t^3+5t^2` `<=>5t^4+10t^3-3t^2-8t-4=0` `<=> (t-1)(t+2)(5t^2+5t+2)=0` `<=>t in {1;-2}` Vs `t=1 <=>(x+1)^2=0<=>x=-1` Vs `t=-2 <=>x^2+2x+2=-2` (vô lí) Vậy `S={-1}` Bình luận
đặt `x^2+2x+3=a`
`⇔1/(a-1)^2+1/a^2=5/4`
`⇔(2a^2-2a+1)/((a^2-2a+1)a^2)=5/4`
`⇔8a^2-8a+4=5a^4-10a^3+5a^2`
`⇔-5x^4-10a^3+3a^2+8a+4=0`
`⇔(1-a)(a+2)(5a^2+5a+2)=0`
`⇔`\(\left[ \begin{array}{l}a=1\\a=-2\end{array} \right.\) `(vì 5a^2+5a+2>0)`
`⇒x=-1`
Đặt `x^2+2x+2=t`
PT `<=>1/t^2 + 1/(t+1)^2=5/4`
`<=>1/t^2 + 1/(t^2+2t+1)=5/4`
`<=>(2t^2+2t+1)/(t^2(t^2+2t+1))=5/4`
`<=>8t^2+8t+4=5t^4+10t^3+5t^2`
`<=>5t^4+10t^3-3t^2-8t-4=0`
`<=> (t-1)(t+2)(5t^2+5t+2)=0`
`<=>t in {1;-2}`
Vs `t=1 <=>(x+1)^2=0<=>x=-1`
Vs `t=-2 <=>x^2+2x+2=-2` (vô lí)
Vậy `S={-1}`