Bài 5:
a: \(\left(\sqrt{5}+2\right)^2=5+2\cdot2\cdot\sqrt{5}+4=9+4\sqrt{5}\)
b: \(\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}\)
\(=\sqrt{5}-2\)
c: \(\left(4-\sqrt{7}\right)^2=16-2\cdot4\cdot\sqrt{7}+7\)
\(=23-8\sqrt{7}\)
d: \(\sqrt{23+8\sqrt{7}}\)
\(=\sqrt{16+2\cdot4\cdot\sqrt{7}+7}\)
\(=4+\sqrt{7}\)
Bài 4:
a: Ta có: \(\sqrt{9x^2}=\left|-9\right|\)
\(\Leftrightarrow3\left|x\right|=9\)
\(\Leftrightarrow\left|x\right|=3\)
hay \(x\in\left\{3;-3\right\}\)
b: Ta có: \(\sqrt{x^4}=9\)
\(\Leftrightarrow x^2=9\)
hay \(x\in\left\{3;-3\right\}\)
c: Ta có: \(\sqrt{9x^2}=2x+1\)
\(\Leftrightarrow\left|3x\right|=2x+1\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=2x+1\left(x\ge0\right)\\3x=-2x-1\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{1}{5}\left(nhận\right)\end{matrix}\right.\)
d: Ta có: \(\sqrt{4x^2-4x+1}=5\)
\(\Leftrightarrow\left|2x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
e: Ta có: \(\sqrt{x^2+6x+9}=3x-1\)
\(\Leftrightarrow\left|x+3\right|=3x-1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=3x-1\left(x\ge-3\right)\\x+3=1-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)
f: Ta có: \(\sqrt{x^2+4x+4}-2=x\)
\(\Leftrightarrow\left|x+2\right|=x+2\)
\(\Leftrightarrow x+2\ge0\)
hay \(x\ge-2\)
