ta có: \(\left(x+4\right)\left(x+1\right)-3\sqrt{x^2+5x+6}=0\)(đk: \(x^2+5x+6\ge0\))
\(\Leftrightarrow x^2+5x+4-3\sqrt{x^2+5x+6}=0\)(1)
đặt \(\sqrt{x^2+5x+5}=a\left(a\ge0\right)\)
khi đó:
(1) \(\Leftrightarrow a^2-1-3\left(a+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(a+1\right)-3\left(a+1\right)=0\)
\(\Leftrightarrow\left(a+1\right)\left(a-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-1\left(loại\right)\\a=4\left(tm\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x^2+5x+5}=4\)
\(\Leftrightarrow x^2+5x-11=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-5+\sqrt{69}}{2}\\x=\frac{-5-\sqrt{69}}{2}\end{matrix}\right.\)(tm)
vậy ...
đề bài thiếu à bạn
