Question

Cho biểu thức A = \(\dfrac{1}{2+\sqrt{x}}\) + \(\dfrac{1}{2-\sqrt{x}}\) - \(\dfrac{2\sqrt{x}}{4-x}\) ; Rút gọn biểu thức A và tìm giá trị của x để A = \(\dfrac{1}{4}\) .

Answer 1

đkxđ: x≥0; x≠4

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

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

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

+) A = 1/4 <=> \(\dfrac{2}{2+\sqrt{x}}=\dfrac{1}{4}\Leftrightarrow2+\sqrt{x}=8\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\)(tm)

Vậy x = 36

Answer 2

đkxđ \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

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

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

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

để \(A=\dfrac{1}{4}\)

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

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

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

vậy tại x=36 thì A=1/4