a: \(A=\left(\dfrac{\left(x-4\right)\left(\sqrt{x}+2\right)-x\sqrt{x}+8}{x-4}\right):\dfrac{x-2\sqrt{x}+4}{\sqrt{x}+2}\)

\(=\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8-x\sqrt{x}+8}{x-4}\cdot\dfrac{\sqrt{x}+2}{x-2\sqrt{x}+4}\)

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

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

\(=\dfrac{-x+4\sqrt{x}-4}{x-2\sqrt{x}+4}=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-1\right)^2+3}< 0\)

=>A<1

c: \(2\sqrt{x}>=0;x-2\sqrt{x}+4=\left(\sqrt{x}-1\right)^2+3>0\)

=>A>=0 với mọi x thỏa mãn  ĐKXĐ

mà A<1

nên 0<=A<1

=>Để A nguyên thì A=0

=>x=0

`1. P = x/(sqrt x-1)`

`= (x-1+1)/(sqrtx-1)`

`= ((sqrt x+1)(sqrt x-1))/(sqrt x-1) +1/(sqrt x-1)`

`= sqrt x+1 + 1/(sqrt x-1)`

`= sqrtx-1 + 1/(sqrt x-1) + 2 >= 4`.

ĐTXR `<=> (sqrtx-1)^2 = 1`.

`<=> x =4` hoặc `x = 0 ( ktm)`.

Vậy Min A `= 4 <=> x= 4`.

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

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

Với x>1\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-1>0\\\dfrac{1}{\sqrt{x}-1}>0\end{matrix}\right.\)

Áp dụng BĐT AM-GM cho 2 số dương \(\sqrt{x}-1\) và \(\dfrac{1}{\sqrt{x}-1}\), ta có:

\(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\ge2\sqrt{(\sqrt{x}-1).\dfrac{1}{\sqrt{x}-1}}=2\)

\(\Rightarrow P\ge2+2=4\)

Dấu = xảy ra khi: \(\sqrt{x}-1=1\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)

KL;....

2:

\(B=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}\)

\(=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\)

=>\(B>=2\cdot\sqrt{25}-6=4\)

Dấu = xảy ra khi (căn x+3)^2=25

=>căn x+3=5

=>căn x=2

=>x=4

mng giúp e với ;-;

4) Ta có: \(P=\dfrac{x-2}{x+2\sqrt{x}}-\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\)

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

\(=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

5) Ta có: \(B=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)

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

=1-x

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

Áp dụng bất đẳng thức Cosi ta có:

\(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\ge2\)

\(\Rightarrow T\ge\sqrt{3\cdot2+6}=2\sqrt{3}\)

Dấu = xảy ra khi x=4

a: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)

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

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

b: P>=-1/2

=>P+1/2>=0

=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)

=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)

=>căn x-9>=0

=>x>=81

c: căn x+3>=3

=>6/căn x+3<=6/3=2

=>-6/căn x+3>=-2

Dấu = xảy ra khi x=0

\(a,A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\left(dk:x\ge0,x\ne4\right)\\ =\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+10-x}{\sqrt{x}+2}\right)\\ =\dfrac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x-4+10-x}\)

\(=\dfrac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\sqrt{x}-2}.\dfrac{1}{6}\\ =\dfrac{-6}{\left(\sqrt{x}-2\right).6}\\ =-\dfrac{1}{\sqrt{x}-2}\)
\(b,A>0\Leftrightarrow-\dfrac{1}{\sqrt{x}-2}>0\Leftrightarrow\sqrt{x}-2< 0\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
Kết hợp với \(dk:x\ge0,x\ne4\), ta kết luận \(0\le x< 4\)

Mình cần gấp nhớ đừng làm tắt nhé 

A = [√x/(x - 4) + 2/(2 - √x) + 1/(√x + 2)] : [(√x - 2 + (10 - x)/(√x + 2)]

= [√x/(√x - 2)(√x + 2) - 2(√x + 2)/(√x - 2)(√x + 2) + (√x - 2)/(√x - 2)(√x + 2)] : [(x - 4 + 10 - x)/(√x + 2)]

= [√x - 2(√x + 2) + (√x - 2)]/[(√x - 2)(√x + 2)] : 6/(√x + 2)

= (√x - 2√x - 4 + √x - 2)/(√x - 2)(√x + 2)] . (√x + 2)/6

= -1/(√x - 2)

Để A > 0 thì -1/(√x - 2) > 0

√x - 2 < 0

√x < 2

x < 4

Vậy 0 ≤ x < 4 thì A > 0