a)
\(\sqrt{x^2+6x+9}=3x-1\)
\(\Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\)
\(\Leftrightarrow\)\(\left|x+3\right|=3x-1\)
\(\Leftrightarrow x+3=3x-1\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
b)
\(\sqrt{1-4x+4x^2}=5\)
\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\)
\(\Leftrightarrow\left|1-2x\right|=5\)
Với \(x\ge\frac{1}{2}\Rightarrow1-2x=5\Leftrightarrow x=-2\)
Với \(x\le\frac{1}{2}\Rightarrow-1+2x=5\Leftrightarrow x=3\)
Tìm x, biết :
a)
\(\sqrt{\left(x+3\right)^2}\) = 3x - 1
\(\Leftrightarrow\) x + 3 = 3x - 1 \(\Leftrightarrow\) x - 3x = - 1 - 3 \(\Leftrightarrow\) x = 2
hoặc x + 3 = -(3x - 1) \(\Leftrightarrow\) x + 3x = 1 - 3 \(\Leftrightarrow\) x = \(\frac{-1}{2}\)
Vậy: S = \(\left\{\frac{-1}{2};2\right\}\)
b)
\(\sqrt{\left(1-2x\right)^2}\) = 5
\(\Leftrightarrow\) (1 - 2x)2 = 52 =25
\(\Leftrightarrow\) 1 - 2x = 5 \(\Leftrightarrow\) x = -2
hoặc 1 - 2x = -5 \(\Leftrightarrow\) x = 3
Vậy: S = \(\left\{-2;3\right\}\)
