Question

[Đặng Hà Minh Huyền]

Cho biểu thức: A=\(\dfrac{1}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-\sqrt{x}}\)(với x≥0,x≠4)

a) Rút gọn A

b) Tìm x để A=\(\dfrac{1}{4}\)

Answer 1

[Huyền]

a, \(A=\dfrac{1}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\left(đkxđ:x\ge0,x\ne4\right)\)

\(A=\dfrac{2-\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{2+\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}-\dfrac{2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)

\(A=\dfrac{2-\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)

\(A=\dfrac{-2\sqrt{x}+4}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)

\(A=\dfrac{2\left(-\sqrt{x}+2\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)

\(A=\dfrac{2}{\sqrt{x}+2}\)

Answer 2

[Huyền]

b, \(A=\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{1}{4}\)

\(\Rightarrow\sqrt{x}+2=8\)

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

\(\Leftrightarrow x=36\left(tm\right)\)

Vậy x = 36 thì \(A=\dfrac{1}{4}\)