ĐK : \(x\in R\)
Đặt : \(x^2+x=t\) ( \(t>0\) )thì phương trình có dạng :
\(\dfrac{12}{t+4}-\dfrac{3}{t+3}=1\)
\(\Leftrightarrow12\left(t+3\right)-3\left(t+4\right)=\left(t+4\right)\left(t+3\right)\)
\(\Leftrightarrow12t+36-3t-12=t^2+7t+12\)
\(\Leftrightarrow t^2-2t-12=0\)
\(\Delta=4+48=52\)
\(\Rightarrow\left\{{}\begin{matrix}t_1=\dfrac{2+\sqrt{52}}{2}=1+\sqrt{13}\left(TM\right)\\t_2=\dfrac{2-\sqrt{52}}{2}=1-\sqrt{13}\left(KTM\right)\end{matrix}\right.\)
\(\Leftrightarrow x^2+x=1+\sqrt{13}\)
\(\Leftrightarrow x^2+x-\left(1+\sqrt{13}\right)=0\)
\(\Delta=1+4+4\sqrt{13}=5+4\sqrt{13}\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-1+\sqrt{5+4\sqrt{13}}}{2}\\x_2=\dfrac{-1-\sqrt{5+4\sqrt{13}}}{2}\end{matrix}\right.\)
Chúc bạn học tốt !!
Đặt \(x^2+x=a\)
\(\Leftrightarrow\dfrac{12}{a+4}-\dfrac{3}{a+3}=1\)
Giải như bình thường !
