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![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\sqrt{x}-3\in\left\{1;2;-2;4\right\}\)
\(\Rightarrow x\in\left\{16;25;1;49\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(\text{đ}k\text{x}\text{đ}:x\ge3\right)\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{2\sqrt{x}-9-\left(x-9\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9-2x+4\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5\sqrt{x}-3x+2}{x-5\sqrt{x}+6}\)
__
Để \(M\in Z\) thì \(x-5\sqrt{x}+6\) thuộc ước của \(5\sqrt{x}-3x+2\)
\(\Rightarrow x-5\sqrt{x}+6=-5\sqrt{x}-3x+2\\ \Leftrightarrow x-5\sqrt{x}+6+5\sqrt{x}+3x-2=0\\ \Leftrightarrow4x-4=0\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ge0;x\ne4\)
\(A=\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b. \(x=36\Rightarrow A=\dfrac{\sqrt{36}}{\sqrt{36}-2}=\dfrac{6}{6-2}=\dfrac{3}{2}\)
c. \(A=-\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=-\dfrac{1}{3}\Rightarrow3\sqrt{x}=2-\sqrt{x}\)
\(\Rightarrow4\sqrt{x}=2\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)
d. \(A>0\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
e. \(A=\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}=1+\dfrac{2}{\sqrt{x}-2}\in Z\Rightarrow\sqrt{x}-2=Ư\left(2\right)\)
\(\Rightarrow\sqrt{x}-2=\left\{-2;-1;1;2\right\}\)
\(\Rightarrow\sqrt{x}=\left\{0;1;3;4\right\}\Rightarrow x=\left\{0;1;9;16\right\}\)
a: Ta có: \(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b: Thay x=36 vào A, ta được:
\(A=\dfrac{6}{6-2}=\dfrac{6}{4}=\dfrac{3}{2}\)
c: Để \(A=-\dfrac{1}{3}\) thì \(3\sqrt{x}=-\sqrt{x}+2\)
\(\Leftrightarrow4\sqrt{x}=2\)
hay \(x=\dfrac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b: Thay \(x=\dfrac{1}{4}\) vào P, ta được:
\(P=\left(\dfrac{1}{2}-1\right):\left(\dfrac{1}{2}+1\right)=\dfrac{-1}{2}:\dfrac{3}{2}=-\dfrac{1}{3}\)
c: Ta có: \(P< \dfrac{1}{2}\)
\(\Leftrightarrow P-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-2-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\sqrt{x}< 3\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm1\end{cases}}\)
Sao phân số thứ 2 là \(\frac{1-2}{1+x}\) .Bạn chép đề thật chuẩn mới trả lời đúng nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
a:
Sửa đề: \(P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{4x^2}{x^2-9}\right):\left(\dfrac{5}{3-x}-\dfrac{4x+2}{3x-x^2}\right)\)\(P=\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{4x^2}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{5x-4x-2}{x\left(3-x\right)}\)
\(=\dfrac{-x^2-6x-9+x^2-6x+9-4x^2}{\left(x-3\right)\left(x+3\right)}:\dfrac{x-2}{x\left(3-x\right)}\)
\(=\dfrac{-4x^2-12x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x\left(3-x\right)}{x-2}\)
\(=\dfrac{-4x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-x\left(x-3\right)}{x-2}=\dfrac{4x^2}{x-2}\)
b: x^2-4x+3=0
=>x=1(nhận) hoặc x=3(loại)
Khi x=1 thì \(P=\dfrac{4\cdot1^2}{1-2}=-4\)
c: P>0
=>x-2>0
=>x>2
d: P nguyên
=>4x^2 chia hết cho x-2
=>4x^2-16+16 chia hết cho x-2
=>x-2 thuộc {1;-1;2;-2;4;-4;8;-8;16;-16}
=>x thuộc {1;4;6;-2;10;-6;18;-14}