\(\sqrt{25-x^2}\) lớn hơn hoặc= 0
=>   25-x² lớn hơn hoặc= 0
=>       -x² lớn hơn hoặc= -25
             x² bé hơn hoặc =25
             x bé hơn hoặc =5
 

a: ĐKXĐ: \(-5\le x\le5\)

b: ĐKXĐ: \(-4\le x\le2\)

mà x nguyên

nên \(x\in\left\{-4;-3;-2;-1;0;1;2\right\}\)

a/ \(đkxđ\) : \(x\ne0;x\ne1\)

b/ 

M = \(\frac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}}{\sqrt{x}-1}-\frac{x+\sqrt{x}}{\sqrt{x}}\)

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

\(=\frac{\left(x-2\sqrt{x}+1\right).\sqrt{x}-\left(x+\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\frac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}+x-x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

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

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

\(=-2\)

chúc bn học tốt

a) ĐKXĐ: \(x\ge0;x\ne9;x\ne4\)

\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(M=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)

\(M=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(M=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(M=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(M=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

b) Ta có M ϵ Z thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3}{\sqrt{x}-3}+\dfrac{4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)

Phải thuộc Z vậy:

4 ⋮ \(\sqrt{x}-3\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Mà: \(x\ge0,x\ne4,x\ne9\) nên \(\sqrt{x}-3\in\left\{1;2;-2;4\right\}\)

\(\Rightarrow x\in\left\{16;25;1;49\right\}\)

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

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

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

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

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

tick cho mình nha

a: ĐKXĐ: x=0; x<>1

\(M=\left(2+\sqrt{x}\right)\left(1-2\sqrt{x}-x+1+\sqrt{x}+x\right)\)

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

b: Sửa đề: P=1/M

P=1/4-x=-1/x-4

Để P nguyên thì x-4 thuộc {1;-1}

=>x thuộc {5;3}

\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(\text{đ}k\text{x}\text{đ}:x\ge3\right)\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{2\sqrt{x}-9-\left(x-9\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9-2x+4\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5\sqrt{x}-3x+2}{x-5\sqrt{x}+6}\)

__

Để \(M\in Z\) thì \(x-5\sqrt{x}+6\) thuộc ước của \(5\sqrt{x}-3x+2\)

\(\Rightarrow x-5\sqrt{x}+6=-5\sqrt{x}-3x+2\\ \Leftrightarrow x-5\sqrt{x}+6+5\sqrt{x}+3x-2=0\\ \Leftrightarrow4x-4=0\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)

a) ĐKXĐ : \(x>0;x\ne1\)

b) \(M=\frac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}}{\sqrt{x}-1}-\frac{x+\sqrt{x}}{\sqrt{x}}\)

\(M=\frac{x+2\sqrt{x}+1-4\sqrt{x}}{\sqrt{x}-1}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\)

\(M=\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}-\left(\sqrt{x}+1\right)\)

\(M=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}-\sqrt{x}-1\)

\(M=\sqrt{x}-1-\sqrt{x}-1\)

\(M=-2\)( đpcm )

\(a,A=\dfrac{2\sqrt{x}-2+2\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\left(x\ge0;x\ne1;x\ne9\right)\\ A=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

\(b,A\in Z\Leftrightarrow\dfrac{\sqrt{x}-3+5}{\sqrt{x}-3}\in Z\Leftrightarrow1+\dfrac{5}{\sqrt{x}-3}\in Z\\ \Leftrightarrow\sqrt{x}-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ Mà.x\ge0\\ \Leftrightarrow\sqrt{x}\in\left\{2;4;8\right\}\\ \Leftrightarrow x\in\left\{4;16;64\right\}\)

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\\x\ne1\end{matrix}\right.\)

\(A=\dfrac{2\sqrt{x}-2+2\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

b) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=1+\dfrac{5}{\sqrt{x}-3}\in Z\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

Kết hợp đk

\(\Rightarrow x\in\left\{4;16;64\right\}\)