Question

cho hai biểu thức A=\(\frac{\sqrt{x}+2}{\sqrt{x}-2}\)và B=\(\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right):\frac{\sqrt{x}+2}{x-4}\)với x≥0,x≠4

a)rút gọn B

b)tìm x nguyên để C=A(B-2)có giá trị nguyên

Answer 1

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

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

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

Để C nguyên \(\Rightarrow\sqrt{x}+2=Ư\left(2\right)=2\)

\(\Rightarrow\sqrt{x}=0\Rightarrow x=0\)

Answer 2

a/ ĐKXĐ:...

\(B=\left(\frac{\sqrt{x}+\sqrt{x}+2}{x-4}\right).\frac{x-4}{\sqrt{x}+2}=\frac{2\sqrt{x}+2}{\sqrt{x}+2}\)

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

\(C=\frac{\sqrt{x}+2}{\sqrt{x}-2}.\frac{2\sqrt{x}+2-2\sqrt{x}-4}{\sqrt{x}+2}=\frac{-2}{\sqrt{x}-2}\)

Để C đạt GT nguyên

\(\Leftrightarrow\sqrt{x}-2\inƯ_{\left(-2\right)}=\left\{\pm2;\pm1\right\}\)

\(\left[{}\begin{matrix}x=0\\x=9\\x=1\\x=16\end{matrix}\right.\)