Question

Tìm x

1,\(\sqrt{x}=2\sqrt{2}\)

2,\(\sqrt{\frac{x+1}{2}}=\frac{\sqrt{5}}{2}\)

Answer 1

1) ĐKXĐ: \(x\ge0\)

\(\sqrt{x}=2\sqrt{2}\Rightarrow x=8\left(tmđkxđ\right)\)

2) ĐKXĐ: \(x\ge-1\)

\(\sqrt{\frac{x+1}{2}}=\frac{\sqrt{5}}{2}\)

\(\Leftrightarrow\frac{x+1}{2}=\frac{5}{4}\)

\(\Leftrightarrow2x+2=5\Leftrightarrow x=\frac{3}{2}\left(TMĐKXĐ\right)\)

Answer 2

1, 

\(\sqrt{x}=2\sqrt{2}\)

=> \(\left(\sqrt{x}\right)=\left(2\sqrt{2}\right)^2\)

=> \(x=8\)

2.

\(\sqrt{\frac{x+1}{2}}=\frac{\sqrt{5}}{2}\)

=> \(\left(\sqrt{\frac{x+1}{2}}\right)=\left(\frac{\sqrt{5}}{2}\right)^2\)

=>  \(\frac{x+1}{2}=\frac{5}{4}\)

=> 4 ( x + 1 ) = 5.2

=> 4x + 4 = 10

=> 4x = 6

=. x = \(\frac{3}{2}\)