Ta có : \(P=3A+2B\)

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

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

Do \(x\ge0\Rightarrow\sqrt{x}+2\ge0\)

\(\Rightarrow-\dfrac{1}{\sqrt{x}+2}\ge-1\)

\(\Rightarrow P=2-\dfrac{1}{\sqrt{x}+2}\ge-1+2=1.\)

Vậy : \(MinP=1.\) Dấu đẳng thức xảy ra khi và chỉ khi \(x=0.\)

`a)` Thay `x=2` vào `B` có: `B=[-10]/[2-4]=5`

`b)` Với `x ne -1;x ne -5` có:

`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+1)(x+5)]`

`A=[x^2+x+2x+2-5x-1-x-5]/[(x+1)(x+5)]`

`A=[x^2-3x-4]/[(x+1)(x+5)]`

`A=[(x+1)(x-4)]/[(x+1)(x+5)]`

`A=[x-4]/[x+5]`

`c)` Với `x ne -5; x ne -1; x ne 4` có:

`P=A.B=[x-4]/[x+5].[-10]/[x-4]`

           `=[-10]/[x+5]`

Để `P` nguyên `<=>[-10]/[x+5] in ZZ`

    `=>x+5 in Ư_{-10}`

Mà `Ư_{-10}={+-1;+-2;+-5;+-10}`

`=>x={-4;-6;-3;-7;0;-10;5;-15}` (t/m đk)

Ta có B=\(\left|x-2\right|+\left|x-4\right|+\left|x-3\right|=\left|x-2\right|+\left|4-x\right|+\left|x-3\right|\ge\left|x-2+4-x\right|+\left|x-3\right|=2+\left|x-3\right|\ge2\)

Dấu = xảy ra <=> x=3

c) Ta có C=\(\left|x-1\right|+\left|4-x\right|+\left|x-2\right|+\left|3-x\right|\ge\left|x-1+4-x\right|+\left|x-2+3-x\right|=4\)

Dấu = xảy ra <=> \(2\le x\le3\)

^_^

b) Ta có: \(\hept{\begin{cases}\left|x-2\right|\ge x-2\\\left|x-3\right|\ge0\\\left|x-4\right|=\left|4-x\right|\ge4-x\end{cases}}\)

\(\Rightarrow\left|x-2\right|+\left|x-3\right|+\left|x-4\right|\ge\left(x-2\right)+\left(4-x\right)\)

\(\Rightarrow B\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2\ge0\\x-3=0\\4-x\ge0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge2\\x=3\\x\le4\end{cases}}\)

Vậy, MinP \(\Leftrightarrow\hept{\begin{cases}x\ge2\\x=3\\x\le4\end{cases}}\)

`B=(x+sqrtx+5)/(sqrtx+1)=(sqrtx(sqrtx+1)+4)/(sqrtx+1)=sqrtx+4/(sqrtx+1)=[(sqrtx+1)+4/(sqrtx+1)]-1>=2\sqrt((sqrtx+1). 4/(sqrtx+1))-1=3`

Dấu "=" xảy ra `<=>x=1`

Vậy `B_(min)=3<=>x=1`