Đáp án + Giải thích các bước giải:
`(x^2+2x+1)/(x^2+2x+2) + (x^2+2x+2)/(x^2+2x+3) = 7/6`
`<=> ((x+1)^2)/((x+1)^2+1) + ((x+1)^2+1)/((x+1)^2+2) = 7/6`
Đặt `(x+1)^2=a`
Phương trình trở thành : `a/(a+1) + (a+1)/(a+2) = 7/6`
`<=> (6a.(a+2)+6(a+1)^2)/(6(a+1).(a+2)) = (7.(a+1).(a+2))/(6.(a+1).(a+2))`
`=> 6a.(a+2)+6(a+1)^2=7.(a+1).(a+2)`
`<=> 6a^2+12a+6.(a^2+2a+1)=7.(a^2+3a+2)`
`<=> 6a^2+12a+6a^2+12a+6=7a^2+21a+14`
`<=> 6a^2+6a^2-7a^2+12a+12a-21a+6-14=0`
`<=> 5a^2+3a-8=0`
`<=> 5a^2-5a+8a-8=0`
`<=> 5a.(a-1)+8.(a-1)=0`
`<=> (5a+8).(a-1)=0`
`<=>` \(\left[ \begin{array}{l}a=-\dfrac{8}{5}\\a=1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}(x+1)^2=-\dfrac{8}{5} \ \ \rm (Loại)\\(x+1)^2=1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x+1=1\\x+1=-1\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\)
Vậy `S={0;-2}`
