a) \(4x^2-1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-1=0\\2x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=-\frac{1}{2}\end{array}\right.\)
b) \(\left(x-1\right)^2=\frac{9}{16}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=\frac{3}{4}\\x-1=-\frac{3}{4}\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{7}{4}\\x=\frac{1}{4}\end{array}\right.\)
c) \(\sqrt{x}=4\left(ĐK:x\ge0\right)\)
\(\Leftrightarrow x=16\)
d) \(\sqrt{x+1}=2\left(ĐKx\ge-1\right)\)
\(\Leftrightarrow x+1=4\)
\(\Leftrightarrow x=3\)
a) \(\Leftrightarrow4x^2=1\Leftrightarrow x^2=\frac{1}{4}\Leftrightarrow x=\frac{1}{2}\)
b) \(\Leftrightarrow x-1=\frac{3}{4}\Leftrightarrow x=\frac{3}{4}+1=\frac{7}{4}\)
c) điều kiện : \(x\ge0\)
\(\sqrt{x}=4\Leftrightarrow x=16\)
d) điều kiện : \(x+1\ge0\Leftrightarrow x\ge-1\)
\(\sqrt{x+1}=2\Leftrightarrow x+1=4\Leftrightarrow x=3\)
