Question

Tìm x, biết :

a)\(\sqrt{x^2+6x+9}=3x-1\)

b)\(\sqrt{1-4x+4x^2}=5\)

Giúp mình với TT

Answer 1

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 \(x\le\frac{1}{2}\Rightarrow-1+2x=5\Leftrightarrow x=3\)

Answer 2

Tìm x, biết :

a) $\sqrt{x^2+6x+9}=3x-1$

$\mathrm{\backslash (\backslash Leftrightarrow\backslash )}$ \(\sqrt{\left(x+3\right)^2}\) = 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) $\sqrt{1-4x+4x^2}=5$

$\mathrm{\backslash (\backslash Leftrightarrow\backslash )}$ \(\sqrt{\left(1-2x\right)^2}\) = 5

\(\Leftrightarrow\) (1 - 2x)² = 5² =25

\(\Leftrightarrow\) 1 - 2x = 5 \(\Leftrightarrow\) x = -2

hoặc 1 - 2x = -5 \(\Leftrightarrow\) x = 3

Vậy: S = \(\left\{-2;3\right\}\)

Answer 3