a) \(\sqrt{16}x^2=3x+1\)
\(\Rightarrow4x^2-3x-1=0\)
\(\Rightarrow\left(x-1\right)\left(4x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{4}\end{matrix}\right.\)
b) \(\sqrt{x^2+4x+4}=x-3\left(1\right)\)
\(\Rightarrow\sqrt{\left(x+2\right)^2}=x-3\)
\(\Rightarrow\left|x+2\right|=x-3\)
TH1: \(x\ge-2\)
\(\left(1\right)\Rightarrow x+2=x-3\Rightarrow2=-3\left(vl\right)\)
TH2: \(x< -2\)
\(\left(1\right)\Rightarrow-x-2=x-3\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\)(không thỏa đk)
Vậy \(S=\varnothing\)
c) \(1-6x+9x^2=7\)
\(\Rightarrow\left(1-3x\right)^2=7\)
\(\Rightarrow\left[{}\begin{matrix}1-3x=\sqrt{7}\\1-3x=-\sqrt{7}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{7}}{3}\\x=\dfrac{1+\sqrt{7}}{3}\end{matrix}\right.\)
d) \(\sqrt{x^4}=3\)
\(\Rightarrow x^2=3\)\(\Rightarrow x=\pm\sqrt{3}\)
a: ta có: \(\sqrt{16x^2}=3x+1\)
\(\Leftrightarrow\left|4x\right|=3x+1\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=3x+1\left(x\ge0\right)\\-4x=3x+1\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{1}{7}\left(nhận\right)\end{matrix}\right.\)
b: Ta có: \(\sqrt{x^2+4x+4}=x-3\)
\(\Leftrightarrow\left|x+2\right|=x-3\)
\(\Leftrightarrow x+2=3-x\left(x\ge-2\right)\)
\(\Leftrightarrow2x=1\)
hay \(x=\dfrac{1}{2}\left(nhận\right)\)
