e)\(=\sqrt{\left(x+4\right)^2}=12\) d)\(\sqrt{\left(x-1\right)^2}=\sqrt{\left(2x-1\right)^2}\)
⇒x+4=12 ⇒x-1=2x-1
⇒x=8 ⇒x=0
\(c.\sqrt{\left(x+4\right)^2}=12\)
\(x+4=12\)
\(x=8\)
\(d.\sqrt{\left(x-1\right)^2}=\sqrt{\left(2x-1\right)2}\)
\(x-1=2x-1\)
\(3x=0\)
\(x=0\)
c) \(\sqrt{x^2+8x+16}=12\)
\(\Leftrightarrow\sqrt[]{\left(x+4\right)^2}=12\)
\(\Leftrightarrow\left|x+4\right|=12\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=12\left(x\ge-4\right)\\x+4=-12\left(x< -4\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12-4=8\left(tm\right)\\x=-12-4=-16\left(tm\right)\end{matrix}\right.\)
d) \(\sqrt{x^2-2x+1}=\sqrt{4x^2-4x+1}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}=\sqrt{\left(2x-1\right)^2}\)
\(\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
\(\sqrt{x^2+8x+16}=12\) ĐKXĐ: x \(\ge-4\)
<=> \(\sqrt{\left(x+4\right)^2}=12\)
<=> x + 4 = 12
<=> x = 12 - 4
<=> x = 8
