\(a,ĐK:x>0;x\ne1\\ b,B=\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ c,B=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\in Z\\ \Leftrightarrow\sqrt{x}-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{2;3\right\}\left(x>0\right)\Leftrightarrow x\in\left\{4;9\right\}\left(tm\right)\)

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

ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

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

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

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

=> Với mọi \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)thì P = 2

Đề sai à --

kkk. thế mới hỏi chứ. đề đấy: đố giải được

\(B=\frac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}.\frac{x-1}{x-2\sqrt{x}}\)

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

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

a.Ta co:

\(\frac{\sqrt{x}-3}{\sqrt{x}-2}< 1\left(x\ge0,x\ne4\right)\) 

\(\Leftrightarrow\sqrt{x}-3< \sqrt{x}-2\)

\(\Leftrightarrow3>2\)

Vay \(B< 1\left(\forall x\ge0,x\ne4\right)\)

Lát mình giải 2 câu kia,di ăn com cái

b.Ta co:

\(\frac{\sqrt{x}-3}{\sqrt{x}-2}< \frac{3}{2}\)

\(\Leftrightarrow2\sqrt{x}-6< 3\sqrt{x}-6\)

\(\Leftrightarrow x>0\)

Vay \(B< \frac{3}{2}\left(\forall x>0,x\ne4\right)\)

c.Ta co:

\(\frac{\sqrt{x}-3}{\sqrt{x}-2}>\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}-3>x-3\sqrt{x}+2\)

\(\Leftrightarrow x-4\sqrt{x}+5< 0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+1< 0\) (vo ly)

Vay khong co gia tri nao cua x thoa man \(B>\sqrt{x}-1\)

(a) Với \(x\ge0,x\ne4\), ta có: 

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

Để \(A\le5\Rightarrow2\sqrt{x}+1\le5\)

\(\Leftrightarrow2\sqrt{x}\le4\Leftrightarrow\sqrt{x}\le2\Leftrightarrow0\le x\le4\).

Kết hợp với điều kiện thì: \(0\le x< 4.\)

(b) \(\dfrac{A}{2}=\dfrac{2\sqrt{x}+1}{2}\) nguyên khi \(\left(2\sqrt{x}+1\right)\in B\left(2\right)=\left\{0;2;4;...;2n\right\}\left(n\in N\right)\)

\(\Leftrightarrow\sqrt{x}\in\left\{-\dfrac{1}{2};\dfrac{1}{2};\dfrac{3}{2};...;\dfrac{2n+1}{2}\right\}\left(n\in N\right)\)

Hay: \(\sqrt{x}\in\left\{\dfrac{1}{2};\dfrac{3}{2};...;\dfrac{2n+1}{2}\right\}\)

\(\Leftrightarrow x\in\left\{\dfrac{1}{4};\dfrac{9}{4};...;\dfrac{\left(2n+1\right)^2}{4}\right\}\)

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

Để A nguyên thì 4 ⋮ √x - 2

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

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

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

Mà x \(\sqrt{x}\ge0\)

=> x thuộc {9; 1; 16; 0; 36}

b)