a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b: Ta có: \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

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

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

c: Ta có: \(x+\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ

\(\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}>0\forall x\)

a: Ta có: \(A=\left(1-\dfrac{2\sqrt{x}-2}{x-1}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{x\sqrt{x}+1}\right)\)

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

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

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

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

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

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

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

b: Thay x=16 vào A, ta được:

\(A=\dfrac{3}{4+3}=\dfrac{3}{7}\)

c)\(A=\dfrac{3}{\sqrt{x}+3}=\dfrac{1}{3}\)

\(\Rightarrow\sqrt{x}+3=9\\ \Rightarrow\sqrt{x}=6\\ \Rightarrow x=36\)

d) \(A=\dfrac{3}{\sqrt{x}+3}\)

Vì \(3>0;\sqrt{x}+3>0\Rightarrow\dfrac{3}{\sqrt{x}+3}>0\)

e) \(2A\in Z\Rightarrow\dfrac{6}{\sqrt{x}+3}\in Z \Rightarrow6⋮x+3\\\Rightarrow\sqrt{x}+3\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\Rightarrow x=\left\{0;9\right\}\)

Biểu thức này không có GTLN vì nếu cho x > 1 và \(x\rightarrow1\Rightarrow\dfrac{1}{\sqrt{x}-1}\rightarrow\infty\).

\(A\le\sqrt{\left(3^2+4^2\right)\left(x-1\right)\left(5-x\right)}=10\)

\(A_{max}=10\) khi \(\dfrac{\sqrt{x-1}}{3}=\dfrac{\sqrt{5-x}}{4}\Rightarrow x=\dfrac{61}{25}\)

\(A=3\left(\sqrt{x-1}+\sqrt{5-x}\right)+\sqrt{5-x}\ge3\left(\sqrt{x-1}+\sqrt{5-x}\right)\ge3\sqrt{x-1+5-x}=6\)

\(A_{min}=6\) khi \(x=5\)

\(M=\left[\frac{\sqrt{x}\left(2\sqrt{x}+3\right)}{2x+2\sqrt{x}+3\sqrt{x}+3}+\frac{2}{\sqrt{x}+1}\right].\frac{\sqrt{x}+2018}{\sqrt{x}+2}\)

\(=\left[\frac{\sqrt{x}\left(2\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+3\right)}+\frac{2}{\sqrt{x}+1}\right].\frac{\sqrt{x}+2018}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}+2}{\sqrt{x}+1}.\frac{\sqrt{x}+2018}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}+2018}{\sqrt{x}+1}\)

\(\frac{\sqrt{x}+2018}{\sqrt{x}+1}=1+\frac{2017}{\sqrt{x}+1}\le2018\)

Dấu "=" xảy ra \(\Leftrightarrow\)

... 

*Max
Xét `P-4`
`=(4\sqrtx+3-4x-4)/(x+1)`
`=(-4x+4\sqrtx-1)/(x+1)`
`=(-(2\sqrtx-1)^2)/(x+1)<=0`
`=>P<=1`
Dấu "=" `<=>2\sqrtx=1<=>x=1/4`
*Min
Xét `P+1`
`=(4\sqrtx+3+x+1)/(x+1)`
`=(x+4\sqrtx+4)/(x+1)`
`=(\sqrtx+2)^2/(x+1)>=0`
`=>P>=-1`
Dấu "=" `<=>\sqrtx+2=0<=>\sqrtx=-2`(vô lý)
=>Không có giá trị nhỏ nhất.

Ta có: \(x+y+z=1\Rightarrow\hept{\begin{cases}\sqrt{x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\\\sqrt{y+xz}=\sqrt{y\left(x+y+z\right)+xz}=\sqrt{\left(x+y\right)\left(y+z\right)}\\\sqrt{z+xy}=\sqrt{z\left(x+y+z\right)+xy}=\sqrt{\left(x+z\right)\left(y+z\right)}\end{cases}}\)

Ta viết lại A

\(A=\sqrt{\left(x+y\right)\left(x+z\right)}+\sqrt{\left(x+y\right)\left(y+z\right)}+\sqrt{\left(y+z\right)\left(x+z\right)}\)

Áp dụng bđt AM-GM:

\(A\le\frac{x+y+x+z+x+y+y+z+y+z+x+z}{2}=2\)

\("="\Leftrightarrow x=y=z=\frac{1}{3}\)

\(x+yz=x\left(x+y+z\right)+yz\)

\(=x^2+xy+xz+yz\)

\(=x\left(x+y\right)+z\left(x+y\right)=\left(x+z\right)\left(x+y\right)\)

+ Tương tự : \(y+xz=\left(x+y\right)\left(y+z\right)\)

\(z+xy=\left(x+z\right)\left(y+z\right)\)

+ Theo bđt AM-GM : \(\sqrt{\left(x+y\right)\left(x+z\right)}\le\frac{x+y+x+z}{2}\)

\(\Rightarrow\sqrt{\left(x-1\right)\left(y-1\right)}\le\frac{2x+y+z}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x+y=x+z\Leftrightarrow y=z\)

+ Tương tự ta cm đc : 

\(\sqrt{\left(x+y\right)\left(y+z\right)}\le\frac{x+2y+z}{2}\).   Dấu "=" xảy ra \(\Leftrightarrow x=z\)

\(\sqrt{\left(x+z\right)\left(y+z\right)}\le\frac{x+y+2z}{2}\).   Dấu "=" xảy ra \(\Leftrightarrow x=y\)

Do đó : \(A\le\frac{4\left(x+y+z\right)}{2}=2\)

A = 2 \(\Leftrightarrow x=y=z=\frac{1}{3}\)

Vậy Max A = 2 \(\Leftrightarrow x=y=z=\frac{1}{3}\)

x+yz=x(x+y+z)+yz

=x2+xy+xz+yz

=x(x+y)+z(x+y)=(x+z)(x+y)

+ Tương tự : y+xz=(x+y)(y+z)

z+xy=(x+z)(y+z)

+ Theo bđt AM-GM : √(x+y)(x+z)≤x+y+x+z2 

⇒√(x−1)(y−1)≤2x+y+z2 

Dấu "=" xảy ra ⇔x+y=x+z⇔y=z

+ Tương tự ta cm đc : 

√(x+y)(y+z)≤x+2y+z2 .   Dấu "=" xảy ra ⇔x=z

√(x+z)(y+z)≤x+y+2z2 .   Dấu "=" xảy ra ⇔x=y

Do đó : A≤4(x+y+z)2 =2

A = 2 ⇔x=y=z=13 

Vậy Max A = 2 

không biết :))))