a: ĐKXĐ: x>0; x<>9
\(D=\dfrac{x+3+\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}}{\sqrt{x}}\cdot\dfrac{1}{\sqrt{x}+3}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
b: x=2+căn 2-căn 2-1=1
Khi x=1 thì \(D=\dfrac{1+1}{1+3}=\dfrac{2}{4}=\dfrac{1}{2}\)
c: P=1:D=(căn x+3)/(căn x+1)
P nguyên
=>căn x+1+2 chia hết cho căn x+1
=>căn x+1 thuộc {1;2}
=>x=0(loại) hoặc x=1(nhận)
a) D có nghĩa khi: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
\(D=\left(\dfrac{x+3}{x-9}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
\(D=\left(\dfrac{x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
\(D=\left(\dfrac{x+3+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
\(D=\dfrac{\left(x+\sqrt{x}\right)\cdot\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(D=\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{x+\sqrt{x}}{x+3\sqrt{x}}\)
\(D=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
b) \(x=\sqrt{6+4\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
\(x=\sqrt{2^2+2\cdot2\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}\)
\(x=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(x=\left|2+\sqrt{2}\right|-\left|\sqrt{2}+1\right|\)
\(x=2+\sqrt{2}-\sqrt{2}-1=1\)
Thay \(x=1\) vào D ta được:
\(D=\dfrac{\sqrt{1}+1}{\sqrt{1}+3}=\dfrac{2}{4}=\dfrac{1}{2}\)
c) Cho \(A=\dfrac{1}{D}=1:\left(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\right)=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1+2}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}+1}+\dfrac{2}{\sqrt{x}+1}=1+\dfrac{2}{\sqrt{x}+1}\)
A nguyên khi \(2⋮\sqrt{x}+1\)
\(\Rightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{1;2;-1;-2\right\}\)
Mà: \(\sqrt{x}>0\)
\(\Rightarrow\sqrt{x}+1\in\left\{1;2\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
