b, \(16x^2-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-x-1\right)\left(4x+x+1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(5x+1\right)=0\)
\(\Rightarrow\) 3x - 1 = 0 hay 5x + 1 = 0
\(3x=1\) \(5x=-1\)
\(x=\dfrac{1}{3}\) \(x=-\dfrac{1}{5}\)
Vậy \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{5}\right\}\)
\(16x^2-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x+x+1\right)\left(4x-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+x+1=0\Leftrightarrow5x=-1\Leftrightarrow x=-\dfrac{1}{5}\\4x-x-1=0\Leftrightarrow3x=1\Leftrightarrow x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{5};\dfrac{1}{3}\right\}\)
b) Ta có: \(16x^2-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-x-1\right)\left(4x+x+1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
