a) \(\sqrt{16x}=8\) b) \(\sqrt{4x}=\sqrt{5}\)
\(\Leftrightarrow16x=8^2\) \(\Leftrightarrow4x=5\)
\(\Leftrightarrow x=\dfrac{8^2}{16}=4\) \(\Leftrightarrow x=\dfrac{5}{4}\)
c) \(\sqrt{9\left(x-1\right)}=21\) c) \(\sqrt{4\left(1-x\right)^2}-6=0\)
\(\Leftrightarrow\sqrt{9\left(x-1\right)=21^2}\) \(\Leftrightarrow\sqrt{2^2.\left(1-x\right)^2}-6=0\)
\(\Leftrightarrow x-1=\dfrac{21^2}{9}\) \(\Leftrightarrow2\left(1-x\right)-6=0\)
\(\Leftrightarrow x-1=49\) \(\Leftrightarrow2-2x-6=0\)
\(\Leftrightarrow x=49+1=50\) \(\Leftrightarrow2x=2-6=-4\)
\(\Leftrightarrow x=-\dfrac{4}{2}=-2\)
a) Ta có: \(\sqrt{16x}=8\)
nên 16x=64
hay x=4
b) Ta có: \(\sqrt{4x}=\sqrt{5}\)
nên 4x=5
hay \(x=\dfrac{5}{4}\)
c) Ta có: \(\sqrt{9\left(x-1\right)}=21\)
\(\Leftrightarrow3\sqrt{x-1}=21\)
\(\Leftrightarrow\sqrt{x-1}=7\)
\(\Leftrightarrow x-1=49\)
hay x=50
d) Ta có: \(\sqrt{4\left(1-x\right)^2}-6=0\)
\(\Leftrightarrow2\left|x-1\right|=6\)
\(\Leftrightarrow\left|x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
