Question

Answer 1

a: Ta có: \(\sqrt{\left(x+2\right)^2}=4\)

\(\Leftrightarrow\left|x+2\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

b: Ta có: \(\sqrt{x^2+10x+25}=1\)

\(\Leftrightarrow\left|x+5\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=1\\x+5=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-6\end{matrix}\right.\)

Answer 2

a. \(\sqrt{\left(x+2\right)^2}=4\)

<=> x + 2 = 4

<=> x = 4 - 2

<=> x = 2

c. \(\sqrt{x^2-2x+1}=3\)

<=> \(\sqrt{\left(x-1\right)^2}=3\)

<=> x - 1 = 3

<=> x = 3 + 1

<=> x = 4

b. \(\sqrt{x^2+10x+25}=1\)

<=> \(\sqrt{\left(x+5\right)^2}=1\)

<=> x + 5 = 1

<=> x = 1 - 5

<=> x = -4

d. \(2\sqrt{x^4=6}\)

<=> 2x² = 6

<=> x² = 3

<=> x = \(\sqrt{3}\)