Question

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

a rút gọn

b. tính giá trị A khi x=25

c. tìm x để A=\(-\dfrac{1}{3}\)

Answer 1

a) \(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}\) ; ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

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

b) Khi x = 25 (t/m) :

\(A=\dfrac{25+4}{25-4}=\dfrac{29}{21}\)

c) \(A=-\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{x+4}{x-4}=-\dfrac{1}{3}\)

\(\Leftrightarrow3\left(x+4\right)=-1\left(x-4\right)\)

\(\Leftrightarrow3x+12=4-x\)

\(\Leftrightarrow4x=-8\)

\(\Leftrightarrow x=-2\) (t/m)

Answer 2

Câu a :

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

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

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

\(=\dfrac{x+4}{x-4}\)

Câu b :

Thay \(x=25\) vào biểu thức A vừa rút gọn ta được :

\(A=\dfrac{25+4}{25-4}=\dfrac{29}{21}\)

Câu c :

\(A=-\dfrac{1}{3}\)

\(\Leftrightarrow\) \(\dfrac{x+4}{x-4}=-\dfrac{1}{3}\)

\(\Leftrightarrow3x+12=-x+4\)

\(\Leftrightarrow4x=-8\)

\(\Leftrightarrow x=-2\)