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

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

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

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

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

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

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

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

__

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) Để \(M\in Z\)

thì \(\sqrt{a}+2⋮\sqrt{a}-2\)

\(\Rightarrow\left(\sqrt{a}-2\right)+4⋮\sqrt{a}-2\)

mà \(\sqrt{a}-2⋮\sqrt{a}-2\)

\(\Rightarrow4⋮\sqrt{a}-2\)

\(\Rightarrow\sqrt{a}-2\inƯ\left(4\right)\)

\(\Rightarrow\sqrt{a}-2\in\left\{\pm1;\pm2;\pm4\right\}\)

......

b) Tìm ra các trường hợp a là số hữu tỉ ở câu a).

a) Ta có : ĐKXĐ :\(a\ne4;a\ge0\)

\(\dfrac{\sqrt{a}+2}{\sqrt{a}-2}=\dfrac{\sqrt{a}-2+4}{\sqrt{a}-2}=\dfrac{\sqrt{a}-2}{\sqrt{a}-2}+\dfrac{4}{\sqrt{a}-2}=1+\dfrac{4}{\sqrt{a}-2}\)

Mà \(A\in Z\) => \(1+\dfrac{4}{\sqrt{a}-2}\in Z=>\dfrac{4}{\sqrt{a}-2}\in Z=>\sqrt{a}-2\inƯ\left(4\right)\)

=>\(\left[{}\begin{matrix}\sqrt{a}-2=1\\\sqrt{a}-2=-1\\\sqrt{a}-2=4\\\sqrt{a}-2=-4\end{matrix}\right.< =>\left[{}\begin{matrix}a=9\\a=1\\a=36\\Loại\end{matrix}\right.\)

Ta có:

\(\dfrac{\sqrt{a}+2}{\sqrt{a}-2}=\dfrac{\left(\sqrt{a}-2\right)+4}{\sqrt{a}-2}=1+\dfrac{4}{\sqrt{a}-2}\)

Để \(M\in Z\)

Thì \(4⋮\left(\sqrt{a}-2\right)\)

hay \(\sqrt{a}-2\inƯ\left(4\right)=\left\{-1;-2;-4;1;2;4\right\}\)

\(\Rightarrow a=\left\{0\right\}\)

Vậy để M là số nguyên thì a={0}

Không biết đúng không

ĐKXĐ \(x\ge0,x\ne4\)

a) \(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}+1\right)-\left(x+5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-2\sqrt{x}-\sqrt{x}+2-\left(x+\sqrt{x}+3\sqrt{x}+3\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{-x-7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\dfrac{-\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}+6}{2-\sqrt{x}}\)

b) B > -1 <=> B + 1 > 0.

\(\Leftrightarrow\dfrac{\sqrt{x}+6}{2-\sqrt{x}}+1>0\Leftrightarrow\dfrac{8}{2-\sqrt{x}}>0\)

=> \(2-\sqrt{x}>0\Leftrightarrow\sqrt{x}< 2\Rightarrow x< 4\)

Vậy \(0\le x< 4\) thì B > -1.

c) \(B=\dfrac{\sqrt{x}+6}{2-\sqrt{x}}=-1-\dfrac{8}{2-\sqrt{x}}\in Z\)

\(\Rightarrow2-\sqrt{x}\inƯ_{\left(8\right)}=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{1;3;0;4;-2;6;-6;10\right\}\)

\(\Rightarrow x\in\left\{1;9;0;16;36;100\right\}\)thì \(B\in Z\)

a) đk : \(x\ne4;x\ge0\)

B = \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}-2}\)

B = \(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-\left(x+5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

B = \(\dfrac{x-2\sqrt{x}-\sqrt{x}+2-\left(x+\sqrt{x}+3\sqrt{x}+3\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

B = \(\dfrac{x-2\sqrt{x}-\sqrt{x}+2-x-\sqrt{x}-3\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

B = \(\dfrac{-x-7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\) = \(\dfrac{\left(-\sqrt{x}-6\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

B = \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\)

a) B = \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}+2}\) ( đk: x \(\ge\) 0; x\(\ne\)4)

<=> B = \(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

<=> B = \(\dfrac{x-3\sqrt{x}+2-x-4\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

<=> B = \(\dfrac{x+7\sqrt{x}+6}{-\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\dfrac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

=> B = \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\)

b) Để B > -1 => \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}>-1\)

=> \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}+1>0\)

<=> \(\dfrac{-8}{\sqrt{x}-2}\) > 0 => \(\sqrt{x}-2< 0\) => \(x< 4\)

Đối chiếu với điều kiện ta được: \(0\le x< 4\)

c) Để B \(\in\) Z => \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\) \(\in\) Z

Mà \(\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\) = \(\dfrac{-\left(\sqrt{x}-2\right)-8}{\sqrt{x}-2}=-1-\dfrac{8}{\sqrt{x}-2}\)

=> \(\dfrac{8}{\sqrt{x}-2}\in Z\) => 8 \(⋮\) \(\sqrt{x}-2\)

=> \(\sqrt{x}\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

=> \(\sqrt{x}\in\left\{3;1;4;0;6;-4;10;-6\right\}\)

Mà \(x\ge0\) => \(x\in\left\{9;0;1;16;36;100\right\}\)

Vậy .......................................................

Ta có: \(M=\dfrac{a^2-3a\sqrt{a}+2}{a-3\sqrt{a}}\)

\(=\dfrac{a^2-a\sqrt{a}-2a\sqrt{a}+2}{a-3\sqrt{a}}\)

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

1: ĐKXĐ: a>=0; a<>1

Đặt \(A=\left(\dfrac{a+2\sqrt{a}}{\sqrt{a}+2}-1\right):\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}+1\right)\)

\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)}{\sqrt{a}+2}-1\right):\left(\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}+1\right)\)

\(=\dfrac{\left(\sqrt{a}-1\right)}{\sqrt{a}+1}\)

2: Để A là số nguyên thì \(\sqrt{a}-1⋮\sqrt{a}+1\)

=>\(\sqrt{a}+1-2⋮\sqrt{a}+1\)

=>\(-2⋮\sqrt{a}+1\)

=>\(\sqrt{a}+1\in\left\{1;-1;2;-2\right\}\)

=>\(\sqrt{a}\in\left\{0;-2;1;-3\right\}\)

=>\(\sqrt{a}\in\left\{0;1\right\}\)

=>\(a\in\left\{0;1\right\}\)

Kết hợp ĐKXĐ, ta được: a=0

a) \(M=3\sqrt{3}-\sqrt{12}-\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(M=3\sqrt{3}-2\sqrt{3}-\left|\sqrt{3}-1\right|\)

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

\(M=1\)

b) Ta có:

\(N=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)

\(N=\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)

\(N=\left(\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)

\(N=\dfrac{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)\cdot\left(\sqrt{a}+1\right)}\)

\(N=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Theo đề ta có: \(M=2N\)

Khi: \(1=2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\)

\(\Leftrightarrow1=\dfrac{2\sqrt{a}-2}{\sqrt{a}}\)

\(\Leftrightarrow\sqrt{a}=2\sqrt{a}-2\)

\(\Leftrightarrow2\sqrt{a}-\sqrt{a}=2\)

\(\Leftrightarrow\sqrt{a}=2\)

\(\Leftrightarrow a=4\left(tm\right)\)