Question

Cho biểu thức N=\(\frac{x}{x-4}\)+\(\frac{1}{\sqrt{x}-2}\)+\(\frac{1}{\sqrt{x}+2}\)

a,Tìm giá trị của N khi x=25

b,Tìm x để N=\(\frac{-1}{3}\)

Answer 1

a)\(N=\frac{x}{x-4}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)

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

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

\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)

Thay x= 25 vào biểu thức N, ta được: \(N=\frac{\sqrt{25}}{\sqrt{25}-2}=\frac{5}{5-2}=\frac{5}{3}\)

Answer 2

b) Để \(N=\frac{-1}{3}\) thì:

\(\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{-1}{3}\)

\(\Leftrightarrow3\sqrt{x}=2-\sqrt{x}\)

\(\Leftrightarrow3\sqrt{x}+\sqrt{x}=2\)

\(\Leftrightarrow4\sqrt{x}=2\)

\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\)

\(\Rightarrow x=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\)