Question

1. Cho biểu thức D = \(\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)

a) tìm giá trị của x để D có nghĩa rồi rút gọn D

b) tìm giá trị của D khi x = \(\sqrt{6+4\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)

c) tìm giá trị của x để \(\dfrac{1}{D}\) nguyên

Answer 1

a)\(ĐK:x\ne9,x\ge0\)

\(D=\left(\dfrac{x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right)\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)

\(=\dfrac{x+3+1\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)

\(=\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

Answer 2

\(x=\sqrt{6+4\sqrt{2}}-\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+2\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}=\left|\sqrt{2}+2\right|-\left|\sqrt{2}+1\right|=\sqrt{2}+2-\sqrt{2}-1=1\)

\(\Rightarrow D=\dfrac{1+1}{1+3}=\dfrac{2}{4}=\dfrac{1}{2}\)

Answer 3

\(\dfrac{1}{D}=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1+2}{\sqrt{x}+1}=1+\dfrac{2}{\sqrt{x}+1}\)

Để \(\dfrac{1}{D}\) nguyên thì \(\left(\sqrt{x}+1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\Leftrightarrow\sqrt{x}\in\left\{0;2;-1;3\right\}\Leftrightarrow x\in\left\{0;4;9\right\}\)