đkxđ: x ∈ R
Đặt: \(x^2+2x+1=t\) (t≥0)
\(pt\Leftrightarrow\sqrt{t+4}+\sqrt{2t+4}=4\)
\(\Leftrightarrow\sqrt{2t+4}=4-\sqrt{t+4}\)
\(\Leftrightarrow2t+4=t+4-8\sqrt{t+4}+16\)
\(\Leftrightarrow t+8\sqrt{t+4}-16=0\)
\(\Leftrightarrow t+4+8\sqrt{t+4}+16-36=0\)
\(\Leftrightarrow\left(\sqrt{t+4}+4\right)^2-36=0\)
\(\Leftrightarrow\left(\sqrt{t+4}+4\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{t+4}+4=6\\\sqrt{t+4}+4=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{t+4}=2\\\sqrt{t+4}=-10\left(loai\right)\end{matrix}\right.\)
Ta có: \(\sqrt{t+4}=2\Leftrightarrow t+4=4\Leftrightarrow t=0\)
Với: t = 0 \(\Leftrightarrow x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\left(tm\right)\)
Vậy pt có một nghiệm x = -1
@Aki Tsuki Tốn thời gian
Lời giải:
\(\sqrt{x^2+2x+5}+\sqrt{2x^2+4x+6}\)
\(=\sqrt{\left(x^2+2x+1\right)+4}+\sqrt{2\left(x^2+2x+1\right)+4}\)
\(=\sqrt{\left(x+1\right)^2+4}+\sqrt{2\left(x+1\right)^2+4}\)
\(\ge\sqrt{4}+\sqrt{4}=2+2=4\)
\("="\Leftrightarrow x=1\)
p/s: Trời vẫn ngát xanh,gió vẫn trong lành :D
