Lời giải:
\(A=\frac{x^2}{\sqrt{x^4+8xy^3}}+\frac{2y^2}{\sqrt{y^4+y(x+y)^3}}\)

Xét:

\(x^4+8xy^3-(x^2+2y^2)^2=8xy^3-4y^4-4x^2y^2\)

\(=-4y^2(x^2-2xy+y^2)=-4y^2(x-y)^2\leq 0\)

\(\Rightarrow x^4+8xy^3\leq (x^2+2y^2)^2\)

\(\Rightarrow \frac{x^2}{\sqrt{x^4+8xy^3}}\geq \frac{x^2}{x^2+2y^2}(*)\)

Mặt khác:
\(y^4+y(x+y)^3-(x^2+2y^2)^2=x^3y+3xy^3-2y^4-x^4-x^2y^2\)

\(=x^3(y-x)+3y^3(x-y)+y^4-x^2y^2\)

\(=x^3(y-x)+3y^3(x-y)+y^2(y-x)(y+x)\)

\(=(y-x)(x^3-2y^3+xy^2)\)

\(=(y-x)[(x-y)(x^2+xy+y^2)+y^2(x-y)]\)

\(=-(x-y)^2(x^2+xy+2y^2)\leq 0\)

\(\Rightarrow y^4+y(x+y)^3\leq (x^2+2y^2)^2\Rightarrow \frac{2y^2}{\sqrt{y^4+y(x+y)^3}}\geq \frac{2y^2}{x^2+2y^2}(**)\)

Từ $(*); (**)\Rightarrow A\geq 1$

a: ĐKXĐ: x>0; x<>1

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

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

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

b: Q>2

=>Q-2>0

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

=>căn x-1>0

=>x>1

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

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

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

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

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

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

b) Q>2 <=> \(\dfrac{x}{\sqrt{x}-1}>2\Leftrightarrow x>2\sqrt{x}-2\)

\(\Leftrightarrow x-2\sqrt{x}+2>0\)

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

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2\ge0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1\le0\\\sqrt{x}-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le1\end{matrix}\right.\)

KL:.....

`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

Biểu thức này ko tồn tại cả min lẫn max

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

Do \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}+2>0\end{matrix}\right.\) \(\Rightarrow\dfrac{\sqrt{x}}{4\left(\sqrt{x}+2\right)}\ge0\)

\(\Rightarrow\dfrac{1}{M}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(x=0\)

ĐKXĐ : \(x\ge2\)

Ta có : \(A=\dfrac{x+3\sqrt{x-2}}{x+4\sqrt{x-2}+1}\) . Đặt t = \(\sqrt{x-2}\ge0\) \(\Rightarrow x=t^2+2\)

Khi đó : \(A=\dfrac{t^2+2+3t}{t^2+4t+3}=\dfrac{\left(t+2\right)\left(t+1\right)}{\left(t+3\right)\left(t+1\right)}=\dfrac{t+2}{t+3}=1-\dfrac{1}{t+3}\ge1-\dfrac{1}{3}=\dfrac{2}{3}\)

" = " \(\Leftrightarrow t=0\Leftrightarrow x=2\)

Vậy ... 

2+ 6/ căn x -1

a: \(B=\dfrac{\sqrt{x}}{x+\sqrt{x}}:\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)

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

\(=\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b: B=2/7

=>\(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{2}{7}\)

=>\(2\left(x+\sqrt{x}+1\right)=7\sqrt{x}\)

=>\(2x+2\sqrt{x}-7\sqrt{x}+2=0\)

=>\(2x-5\sqrt{x}+2=0\)

=>\(\left(2\sqrt{x}-1\right)\cdot\left(\sqrt{x}-2\right)=0\)

=>\(\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)

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

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

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

\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{1}{\sqrt{x}-1}\)

\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)\)

\(=3\sqrt{x}-6\)

b) \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)

\(\Leftrightarrow3\sqrt{x}-6=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)   (1)

ĐKXĐ: \(x>0\)

\(\left(1\right)\Leftrightarrow3x-6\sqrt{x}=4\sqrt{x}-1\)

\(\Leftrightarrow3x-6\sqrt{x}-4\sqrt{x}+1=0\)

\(\Leftrightarrow3x-10\sqrt{x}+1=0\)   (2)

Đặt \(t=\sqrt{x}\ge0\)

\(\left(2\right)\Leftrightarrow3t^2-10t+1=0\)

\(\Delta'=25-4=22\)

Phương trình có hai nghiệm phân biệt:

\(t_1=\dfrac{5+\sqrt{22}}{3}\) (nhận)

\(t_2=\dfrac{5-\sqrt{22}}{3}\) (nhận)

Với \(t=\dfrac{5+\sqrt{22}}{3}\) \(\Leftrightarrow\sqrt{x}=\dfrac{5+\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47+10\sqrt{22}}{9}\) (nhận)

Với \(t=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow\sqrt{x}=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47-10\sqrt{22}}{9}\) (nhận)

Vậy \(x=\dfrac{47+10\sqrt{22}}{9};x=\dfrac{47-10\sqrt{22}}{9}\) thì \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)

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

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

b: P=(4căn x-1)/căn x

=>3x-6căn x-4căn x+1=0

=>3x-10căn x+1=0

=>x=(47+10căn 22)/9 hoặc x=(47-10căn 22)/9

*Rút gọn

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

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

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

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

Ta có: \(C=x-\sqrt{x}+1\)

\(=x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\sqrt{x}=\dfrac{1}{2}\)

hay \(x=\dfrac{1}{4}\)

\(C=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\left(x>0;x\ne1\right)\)

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

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

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

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

Dấu "=" xảy ra khi \(\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}\)

Vậy  \(C_{min}=\dfrac{3}{4}\)

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

Áp dụng AM-GM có: \(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\)

Dấu "=" xảy ra khi x=1 (ktm đk)

Suy ra dấu bằng ko xảy ra \(\Rightarrow\sqrt{x}+\dfrac{1}{\sqrt{x}}-1>2-1=1\)

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

\(\Rightarrow N< 2\) mà \(N>0\),\(N\) nguyên

\(\Rightarrow N=1\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=1\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{3+\sqrt{5}}{2}\\\sqrt{x}=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7+3\sqrt{5}}{2}\\x=\dfrac{7-3\sqrt{5}}{2}\end{matrix}\right.\) (tm)

Vậy...

\(\Rightarrow C=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\) * \(\Rightarrow C=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\) Dấu = xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)

* Ta có \(N=\dfrac{2\sqrt{x}}{C}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}>0\left(1\right)\) 

Xét \(N-2=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}-2=\dfrac{2\sqrt{x}-2x+2\sqrt{x}-2}{x-\sqrt{x}+1}=\dfrac{-2x+4\sqrt{x}-2}{x-\sqrt{x}+1}=\dfrac{-2\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}< 0\left(dox\ne1\right)\Rightarrow N< 2\left(2\right)\) Từ (1) và (2) \(\Rightarrow0< N< 2\). Mà N nguyên nên N=1  \(\Rightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=1\Rightarrow2\sqrt{x}=x-\sqrt{x}+1\Leftrightarrow x-3\sqrt{x}+1=0\)

\(\Delta=9-4=5\Rightarrow\) pt có 2 nghiệm phân biệt: \(x_1=\dfrac{\sqrt{5}+3}{2}\left(TM\right);x_2=\dfrac{3-\sqrt{5}}{2}\left(TM\right)\)