ĐKXĐ: \(x\in\mathbb{R}\)
\(\dfrac{1}{\left(x^2+x+1\right)^2}+\dfrac{1}{\left(x^2+x+2\right)^2}=\dfrac{13}{36}\) (1)
Đặt \(x^2+x+1=t;\left(t>0\right)\), thì (1) trở thành:
\(\dfrac{1}{t^2}+\dfrac{1}{\left(t+1\right)^2}=\dfrac{13}{36}\\ \Leftrightarrow\dfrac{36\left(t+1\right)^2}{36t^2\left(t+1\right)^2}+\dfrac{36t^2}{36t^2\left(t+1\right)^2}=\dfrac{13t^2\left(t+1\right)^2}{36t^2\left(t+1\right)^2}\\ \Rightarrow36\left(t^2+2t+1\right)+36t^2=13t^2\left(t^2+2t+1\right)\\ \Leftrightarrow13t^4+26t^3+13t^2=72t^2+72t+36\\ \Leftrightarrow13t^4+26t^3-59t^2-72t-36=0\\ \Leftrightarrow13t^4-26t^3+52t^3-104t^2+45t^2-90t+18t-36=0\\ \Leftrightarrow13t^3\left(t-2\right)+52t^2\left(t-2\right)+45t\left(t-2\right)+18\left(t-2\right)\\=0\\ \Leftrightarrow\left(t-2\right)\left(13t^3+52t^2+45t+18\right)=0\)
\(\Leftrightarrow\left(t-2\right)\left(13t^3+39t^2+13t^2+39t+6t+18\right)=0\\ \Leftrightarrow\left(t-2\right)\left[13t^2\left(t+3\right)+13t\left(t+3\right)+6\left(t+3\right)\right]=0\\ \Leftrightarrow\left(t-2\right)\left(t+3\right)\left(13t^2+13t+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t-2=0\\t+3=0\\13t^2+13t+6=0\end{matrix}\right.\)
+, Với `t-2=0=>x^2+x+1-2=0`
`<=>x^2+x-1=0`
`<=>[x^2+2.x. 1/2+(1/2)^2]-1/4-1=0`
`<=>(x+1/2)^2-5/4=0`
`<=>(x+{1-\sqrt5}/{2})(x+{1+\sqrt5}/{2})=0`
`<=>[(x={-1+\sqrt5}/{2}),(x={-1-\sqrt5}/{2}):}`
+, Với `t+3=0<=>t=-3` (ktm)
+, Với `13t^2+13t+6=0`
`<=>13(t^2+t)+6=0`
`<=>13[t^2+2.t. 1/2+(1/2)^2]-13/4+6=0`
`<=>13(t+1/2)^2+21/4=0` (vô lí)
Vậy phương trình đã cho có tập nghiệm là: $S=\{\dfrac{-1+\sqrt5}{2};\dfrac{-1-\sqrt5}{2}\}$.
#$\mathtt{Toru}$
