a) ĐKXĐ: \(x\ge0,x\ne1\)

\(P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

b) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=-1\)

\(\Leftrightarrow-\sqrt{x}-1=\sqrt{x}-1\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)

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

\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Kết hợp đk:

\(\Leftrightarrow x\in\left\{0\right\}\)

d) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\)

e) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)

Do \(\sqrt{x}+1\ge1\Leftrightarrow-\dfrac{2}{\sqrt{x}+1}\ge-2\)

\(\Leftrightarrow P=1-\dfrac{2}{\sqrt{x}+1}\ge1-2=-1\)

\(minP=-1\Leftrightarrow x=0\)

\(a,P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\left(x\ge0;x\ne1\right)\\ P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,P=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\\ \Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\\ c,P\in Z\Leftrightarrow\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{1;2\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}=0\left(x\ne1\right)\\ \Leftrightarrow x=0\)

\(d,P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\left(\dfrac{2}{\sqrt{x}+1}>0\right)\\ e,P=1-\dfrac{2}{\sqrt{x}+1}\\ \sqrt{x}+1\ge1\Leftrightarrow-\dfrac{2}{\sqrt{x}+1}\ge-\dfrac{2}{1}=-2\\ \Leftrightarrow P=1-\dfrac{2}{\sqrt{x}+1}\ge1-\left(-2\right)=3\)

Dấu \("="\Leftrightarrow x=0\)

a: R-3=(x^2+x-1-3x)/x=(x-1)^2/x

Nếu x>0 thì R-3>0

=>R>3

Nếu x<0 thì R-3<0

=>R<3

c: Để R>4 thì R-4>0

=>\(\dfrac{x^2+x+1-4x}{x}>0\)

=>\(\dfrac{x^2-3x+1}{x}>0\)

TH1: x>0 và x^2-3x+1>0

=>x>0 và \(\left[{}\begin{matrix}x< \dfrac{3-\sqrt{5}}{2}\\x>\dfrac{3+\sqrt{5}}{2}\end{matrix}\right.\Leftrightarrow x>\dfrac{3+\sqrt{5}}{2}\)

mà x nguyên

nên x>3

TH2: x<0 và x^2-3x+1<0

=>x<0 và \(\dfrac{3-\sqrt{5}}{2}< x< \dfrac{3+\sqrt{5}}{2}\)(loại)

Mk làm từng câu nhé !

a)\(A=\frac{x-\sqrt{x}}{x-1}\left(đk:x\ge0,x\ne1\right)\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)(vì \(x\ge0\))

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

\(B=\frac{x-4}{x+2\sqrt{x}}\left(đk:x>0,x\ne4\right)\)

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

a.\(DK:x\ge0,x\ne1\)

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

\(DK:x\ge0\)

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

b.\(A-B=\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{x-x+\sqrt{x}+2}{x+\sqrt{x}}=\frac{\sqrt{x}+2}{x+\sqrt{x}}>0\) 

\(\Rightarrow A-B>0\Rightarrow A>B\)

c.Ta co:\(A.B=\frac{\sqrt{x}}{\sqrt{x}+1}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{\sqrt{x}-2}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)

De \(A.B\in Z\)

\(\Rightarrow1-\frac{3}{\sqrt{x}+1}\in Z\)

\(\Rightarrow\frac{3}{\sqrt{x}+1}\in Z\)

\(\Rightarrow3⋮\sqrt{x}+1\)

\(\Rightarrow x=4\)

d.Ta co: \(A.B=\frac{\sqrt{x}-2}{\sqrt{x}+1}< \frac{1}{2}\)

\(\Leftrightarrow2\sqrt{x}-4< \sqrt{x}+1\)

\(\Leftrightarrow x< 25\)

Để làm 1 lượt luôn, hihi !

b)

Có \(A=\frac{\sqrt{x}}{\sqrt{x}+1}\)\(< 1\)(vì \(x\ge0\))

\(B=1+\frac{2}{\sqrt{x}}>1\)(Vì \(x\ge0\))

=>A<B

c)Ta có:

A.B=\(\frac{\sqrt{x}}{\sqrt{x}+1}.\frac{2}{\sqrt{x}}=\frac{\sqrt{x}+2}{\sqrt{x}+1}=1+\frac{1}{\sqrt{x}+1}\)

Để:\(A.B\in\frac{Z\Rightarrow1}{\sqrt{x}+1}\in Z\)

Với:\(x\ge0\)có \(\orbr{\begin{cases}\sqrt{x}\in Z\\\sqrt{x}\notin Z\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+1\in Z\\\sqrt{x}+1\notin Z\end{cases}\Leftrightarrow}\orbr{\begin{cases}\frac{1}{\sqrt{x}+1}\in Z\left(tm\right)\\\frac{1}{\sqrt{x}+1}\notin Z\left(ktm\right)\end{cases}}}\)

\(\Rightarrow\sqrt{x}+1\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;-2\right\}\)

Mà \(\sqrt{x}\ge0\Rightarrow\sqrt{x}=0\Leftrightarrow x=0\left(\frac{t}{m}\right)\)

Vậy: để A.B thuộc Z <=> x=0

d)Có A.B < 1/2

<=>\(1+\frac{1}{\sqrt{x}+1}< \frac{1}{2}\Leftrightarrow\frac{1}{\sqrt{x}+1}+\frac{1}{2}< 0\)(vô lý)

Vậy không có x t/m A.B < 1/2