Question

b,,\(\sqrt{x^2+4x+8}+\sqrt{x^2+4x+4}=\sqrt{2\cdot\left(x^2+4x+6\right)}\)

Answer 1

Đặt \(x^2+4x+8=a\)(a > 0 vì \(x^2+4x+8=\left(x+2\right)^2+4>0\)) , \(x^2+4x+4=b\left(b\ge0\right)\)
\(\Rightarrow a+b=x^2+4x+8+x^2+4x+4=2\left(x^2+4x+6\right)\)
Khi đó phương trình đã cho trở thành:
\(\sqrt{a}+\sqrt{b}=\sqrt{a+b}\\ \Leftrightarrow a+b+2\sqrt{ab}=a+b\\ \Leftrightarrow2\sqrt{ab}=0\\\Leftrightarrow\sqrt{ab}=0\\ \Leftrightarrow ab=0\\ \Leftrightarrow\left[{}\begin{matrix}a=0\left(kh\text{ô}ngtm\right)\\b=0\end{matrix}\right.\)
\(\Rightarrow x^2+4x+4=0\\ \Leftrightarrow\left(x+2\right)^2=0\\ \Leftrightarrow x+2=0\\ \Leftrightarrow x=-2.\)

Kl: