Question

Answer 1

a.

ĐKXĐ: \(\dfrac{-5}{2+x}\ge0\)

\(\Leftrightarrow2+x< 0\)

\(\Leftrightarrow x< -2\)

b.

Ta có: \(3\sqrt{2}=\sqrt{3^2.2}=\sqrt{18}\)

\(\dfrac{1}{5}\sqrt{425}=\dfrac{1}{5}\sqrt{5^2.17}=\dfrac{5}{5}\sqrt{17}=\sqrt{17}\)

Mà \(\sqrt{18}>\sqrt{17}\Rightarrow3\sqrt{2}>\dfrac{1}{5}\sqrt{425}\)

Answer 2

a: ĐKXĐ: \(x< -2\)

b: Ta có: \(3\sqrt{2}=\sqrt{18}\)

\(\dfrac{1}{5}\sqrt{425}=\sqrt{425\cdot\dfrac{1}{25}}=\sqrt{17}\)

mà 18>17

nên \(3\sqrt{2}>\dfrac{1}{5}\sqrt{425}\)