Question

[Linh Nguyễn]

Mình cần gấp ạ!!! Cảm ơn!

Answer 1

[Nguyễn Lê Phước Thịnh]

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)

Answer 2

[HT.Phong (9A5)]

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: \(\sqrt{x}>0\) 

\(\Rightarrow\sqrt{x}+1\in\left\{1;2\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)