Question

giải phương trình sau :

\(\dfrac{x\left(x+1\right)\left(x+4\right)\left(x+3\right)+1}{\left(x+2\right)^2\left(x+5\right)\left(x-1\right)+2}=3\)

Answer 1

\(\Leftrightarrow\dfrac{\left(x^2+4x\right)\left(x^2+4x+4\right)+1}{\left(x^2+4x+4\right)\left(x^2+4x-5\right)+2}=3\)

\(\Leftrightarrow\dfrac{\left(x^2+4x\right)^2+4\left(x^2+4x\right)+1}{\left(x^2+4x\right)^2-\left(x^2+4x\right)-18}=3\)

Đặt \(x^2+4x=a\)

Pt sẽ là \(\dfrac{a^2+4a+1}{a^2-a-18}=3\)

=>3a^2-3a-54=a^2+4a+1

=>2a^2-7a-55=0

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{7+\sqrt{489}}{4}\\a=\dfrac{7-\sqrt{489}}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+4x-\dfrac{7+\sqrt{489}}{4}=0\\x^2+4x-\dfrac{7-\sqrt{489}}{4}=0\end{matrix}\right.\)

=>\(x\in\left\{2.07;-6.07;0.34;-4.34\right\}\)