\(\sqrt{3x+11}=3+\sqrt{2}\)
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a) \(x^2+8=3\sqrt{x^3+8}\)
\(\left(x^2+8\right)^2=\left(3\sqrt{x^2+8}\right)^2\)
\(x^4+16x^2+64=9x^2+72\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)
\(\Leftrightarrow\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
Ta có:
\(\sqrt{3x+11}=3+\sqrt{2}\Leftrightarrow\left(\sqrt{3x+11}\right)^2=\left(3+\sqrt{2}\right)^2\)
\(\Leftrightarrow3x+11=9+6\sqrt{2}+2\)
\(\Leftrightarrow3x=6\sqrt{2}\Leftrightarrow x=2\sqrt{2}\)
\(\sqrt{3x+11}=3+\sqrt{2}\Leftrightarrow\left(\sqrt{3x+11}\right)^2=\left(3+\sqrt{2}\right)^2\)
\(\Leftrightarrow3x+11=9+2+6\sqrt{2}\)
\(\Leftrightarrow3x+11=11+6\sqrt{2}\)
\(\Leftrightarrow3x=6\sqrt{2}\Leftrightarrow x=2\sqrt{2}\)