Question

Tìm x²:

a)\(\sqrt{x}=\frac{1}{2}\)

b)\(\sqrt{3x}=9\)

c)\(2\sqrt{x}+1=7\)

d)\(9-5\sqrt{x}=-1\)

Answer 1

a) ĐKXĐ: x≥0

Ta có: \(\sqrt{x}=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{4}\)(nhận)

hay \(x^2=\frac{1}{16}\)

Vậy: \(x^2=\frac{1}{16}\)

b) ĐKXĐ: x≥0

Ta có: \(\sqrt{3x}=9\)

\(\Leftrightarrow3x=81\)

\(\Leftrightarrow x=27\)(nhận)

hay \(x^2=729\)

Vậy: \(x^2=729\)

c) ĐKXĐ: x≥0

Ta có: \(2\sqrt{x}+1=7\)

\(\Leftrightarrow2\sqrt{x}=6\)

\(\Leftrightarrow\sqrt{x}=3\)

\(\Leftrightarrow x=9\)(nhận)

hay \(x^2=81\)

Vậy: \(x^2=81\)

d) ĐKXĐ: x≥0

Ta có: \(9-5\sqrt{x}=-1\)

\(\Leftrightarrow5\sqrt{x}=10\)

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

\(\Leftrightarrow x=4\)(nhận)

hay \(x^2=16\)

Vậy: \(x^2=16\)