ĐKXĐ: \(x\ge0\)

\(Q=\dfrac{3\sqrt{x}+15}{\sqrt{x}+3}=\dfrac{3\left(\sqrt{x}+3\right)+6}{\sqrt{x}+3}=3+\dfrac{6}{\sqrt{x}+3}\)

Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\Rightarrow\dfrac{6}{\sqrt{x}+3}\le2\Rightarrow3+\dfrac{6}{\sqrt{x}+3}\le5\)

\(\Rightarrow Q_{max}=5\) khi \(x=0\)

\(Q-5=\dfrac{3\sqrt{x}+15}{\sqrt{x}+3}-5=\dfrac{-2\sqrt{x}}{\sqrt{x}+3}\le0\)

\(\Rightarrow Q\le5\)

\(Q_{maxx}=5\) khi \(\sqrt{x}=0\Leftrightarrow x=0\)

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

Để P nguyên thì \(2⋮\sqrt{x}-3\Leftrightarrow\sqrt{x}-3\inƯ\left(2\right)=\left\{\pm1,\pm2\right\}\)

\(\begin{matrix}\sqrt{x}-3&-1&-2&1&2\\\sqrt{x}&-2\left(L\right)&1&4&5\\x&&1\left(tm\right)&16\left(tm\right)&25\left(tm\right)\end{matrix}\) 

Mà x nguyên lớn nhất \(\Rightarrow x=25\)

Để P là số nguyên thì

căn x-3-2 chia hết cho căn x-3

=>căn x-3 thuộc Ư(-2)

mà x nguyên lớn nhất

nên căn x-3=2

=>x=25

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

Để \(P \in Z \Leftrightarrow 2\vdots \sqrt x -3 \Rightarrow \sqrt x -3 \in \text{Ư(2)={1;-1;2;-2}}\)

\(\Rightarrow \sqrt x \in \text{{4;2;5;1}} \Rightarrow x \in \text{{16;4;25;1}}\)

\(\Rightarrow x_{max}=25\)

a: \(A=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}=\dfrac{-3\sqrt{x}-9}{x-9}\)

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

b: A=1/3

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

=>căn x-3=-9

=>căn x=-6(loại)

c: căn x-3>=-3

=>3/căn x-3<=-1

=>-3/căn x-3>=1

Dấu = xảy ra khi x=0

`C=(sqrtx+3)/(sqrtx-2)=(sqrtx-2+5)/(sqrtx-2)=1+5/(sqrtx-2)`

Ta cần tìm `max(5/(sqrtx-2))`

Nếu `0<=x<4` thì `5/(sqrtx-2)<0`

Nếu `x>4` thì `5/(sqrtx-2)>0`

Do đó ta chỉ xét `x>4` hay `x>=5(` Do `x` nguyên `)`

`=>sqrtx-2>=sqrt5-2`

`=>5/(sqrtx-2)<=5/(sqrt5-2)`

`=>C<=1+5/(sqrt5-2)=11+sqrt5`

Vậy `C_(max)=11+sqrt5<=>x=5`

`B = (\sqrt{x} + 3)/(\sqrt{x} - 3)`
`=>B = (6 + \sqrt{x} -3)/(\sqrt{x} - 3)`
`=>B = 1 + 6/(\sqrt{x-3})`
Để `B` đạt gt lớn nhất

`=>6 \vdots \sqrt{x-3}`
`=>12 \vdots (x-3)`
`=>(x-3)\in Ư(12) = {+-1;+-2;+-3;+-4;+-6;+-12}`
Do `x` là stn
`=>(x-3) \in {1;2;3;4;6;12}`
`=>x = 15`
Vậy `x=15` 

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

\(ĐK:x\ge0;x\ne1;x\ne-2\)

\(P=\dfrac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\dfrac{3x+3\sqrt{x}-9+x-\sqrt{x}+3\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)

b.\(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2}{\sqrt{x}+2}+\dfrac{\sqrt{x}+2}{\sqrt{x}+2}+\dfrac{\sqrt{x}+2}{\sqrt{x}+2}+\dfrac{2}{\sqrt{x}+2}\)

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

Để P lớn nhất thì \(\sqrt{x}+2\) nhỏ nhất

Mà \(\sqrt{x}+2\ge2\) \(\Rightarrow Min=2\)

\(\Rightarrow P\le1+1+1+\dfrac{2}{2}=1+1+1+1=4\)

Vậy \(P_{max}=4\) khi \(x=0\)

a: \(A=\dfrac{2\sqrt{x}+6+\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{3}{\sqrt{x}+3}\)

b: \(\sqrt{x}+3>=3\)

=>A<=1

Dấu = xảy ra khi x=0

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

Để P nguyên thì căn x-2\(\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{1;25\right\}\)