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

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

__

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

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

\(a.P=\dfrac{2\sqrt{x}-5}{x-5\sqrt{x}+4}+\dfrac{2}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-4}=\dfrac{2\sqrt{x}-5+2\sqrt{x}-8-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)}=\dfrac{3}{\sqrt{x}-1}\) ( x ≥ 0 ; x # 1 ; x # 16 )

\(b.\) \(P\text{∈}Z\) ⇔ \(\dfrac{3}{\sqrt{x}-1}\text{∈}Z\) ⇔ \(\sqrt{x}-1\text{∈}\left\{1;-1;3;-3\right\}\)

+) \(\sqrt{x}-1=1\text{⇔}x=4\left(TM\right)\)

+) \(\sqrt{x}-1=-1\text{⇔}x=0\left(TM\right)\)

+) \(\sqrt{x}-1=3\text{⇔}x=16\left(KTM\right)\)

+) \(\sqrt{x}-1=-3\text{⇔}vo-nghiem\)

KL............

\(B=\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

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

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

Để B nguyên thì \(\sqrt{x}-3\in\left\{1;-1;5\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;2;8\right\}\)

hay \(x\in\left\{16;4;64\right\}\)

1: \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{2}=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)

2: Để P là số nguyên thì \(2\sqrt{x}+2⋮2\sqrt{x}\)

\(\Leftrightarrow2\sqrt{x}=2\)

hay x=1(nhận)

3: \(P-\dfrac{1}{2}=\dfrac{\sqrt{x}+1}{2\sqrt{x}}-\dfrac{1}{2}=\dfrac{2\sqrt{x}+2-\sqrt{x}}{2\sqrt{x}}=\dfrac{\sqrt{x}+2}{2\sqrt{x}}>0\)

=>P>1/2

Xem lại đề bạn ơi

kia đáng phải + chứ

1. a) \(A=\left(\dfrac{\sqrt{x}-1+x-\sqrt{x}}{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)}\right).\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)ĐK x\(\ne\)0,1

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

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

b) A<-1 <=> \(\dfrac{2\sqrt{x}}{x-\sqrt{x}}< -1\)\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}+1< 0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}+x-\sqrt{x}}{x-\sqrt{x}}< 0\)\(\Leftrightarrow\dfrac{x+\sqrt{x}}{x-\sqrt{x}}< 0\)

\(\Leftrightarrow x-\sqrt{x}< 0\) (vì \(x+\sqrt{x}>0\left(\forall x>0\right)\))

\(\Leftrightarrow x< \sqrt{x}\Leftrightarrow x^2< x\Leftrightarrow x^2-x< 0\)

\(\Leftrightarrow x\in\left(0;1\right)\Leftrightarrow0< x< 1\)

đk x khác 0 

\(A=4+\dfrac{6}{\sqrt{x}}\Rightarrow\sqrt{x}\inƯ\left(6\right)=\left\{1;2;3;6\right\}\)

a) đkxđ x≥0 , x ≠1

\(K=\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

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

để K ∈ z thì \(\dfrac{-1}{\sqrt{x}-2}\) nguyên

=> √x -2 ∈ Ư(-1)={-1;1}

=> x ∈ {1; 9}

vậy ...

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

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

b: Để K là số nguyên thì \(\sqrt{x}-2-1⋮\sqrt{x}-2\)

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

hay x=9

c: Để K là số âm thì \(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}< 0\)

=>4<x<9