Question

A=\(\dfrac{\sqrt{x}}{\sqrt{x}-5}+\dfrac{10-\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)

a) Rút gọn.

b) Tìm x để \(A< \dfrac{1}{3}\)

Answer 1

ĐKXĐ:\(x\ne25\)

a) rút gọn A=\(\dfrac{\sqrt{x}}{\sqrt{x}+5}+\)\(\dfrac{10-\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\) = \(\dfrac{\sqrt{x}\left(\sqrt{x}+5\right)+10-\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\) =\(\dfrac{x+5\sqrt{x}+10-\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\) =\(\dfrac{x-\sqrt{x}+35}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)

Answer 2

b) Tìm x để A>\(\dfrac{1}{3}\)

Ta có A=\(\dfrac{x-\sqrt{x}+35}{x-25}< \dfrac{1}{3}\)

\(\Leftrightarrow\)\(3\left(x-\sqrt{x}+35\right)< x-25\)

\(\Leftrightarrow\)\(3x-3\sqrt{x}+105< x-25\)

\(\Leftrightarrow\)\(3x-3\sqrt{x}+105-x+25< 0\)