a) Với \(x\ge0\)và \(x\ne1\)ta có:

\(P=\frac{10\sqrt{x}}{x+3\sqrt{x}-4}-\frac{2\sqrt{x}-3}{\sqrt{x}+4}+\frac{\sqrt{x}+1}{1-\sqrt{x}}\)

\(=\frac{10\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}-\frac{2\sqrt{x}-3}{\sqrt{x}+4}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\frac{10\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}-\frac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)

\(=\frac{10\sqrt{x}-\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)

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

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

\(=\frac{-3x+10\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}=\frac{-\left(3x-10\sqrt{x}+7\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)

\(=\frac{-\left(\sqrt{x}-1\right)\left(3\sqrt{x}-7\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}=\frac{-3\sqrt{x}+7}{\sqrt{x}+4}\)

b) \(P=\frac{-3\sqrt{x}+7}{\sqrt{x}+4}=\frac{-3\sqrt{x}-12+19}{\sqrt{x}+4}=\frac{-3\left(\sqrt{x}+4\right)+19}{\sqrt{x}+4}=-3+\frac{19}{\sqrt{x}+4}\)

Vì \(x\ge0\); \(x\ne1\)\(\Rightarrow\sqrt{x}+4\ge4\)

\(\Rightarrow\frac{19}{\sqrt{x}+4}\le\frac{19}{4}\)\(\Rightarrow P\le-3+\frac{19}{4}=\frac{7}{4}\)

Dấu " = " xảy ra \(\Leftrightarrow x=0\)( thỏa mãn )

Vậy \(maxP=\frac{7}{4}\)\(\Leftrightarrow x=0\)

Câu 1: 

a: \(Q=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

b: Để Q>0 thì \(\sqrt{a}-2>0\)

=>a>4

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

\(A=\left(\dfrac{1}{1-\sqrt{x}}+\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right):\dfrac{\sqrt{x}-1}{3}=\dfrac{-x-\sqrt{x}-1+x+2+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{3}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{3}{\sqrt{x}-1}=\dfrac{3}{x+\sqrt{x}+1}\text{≤}\dfrac{3}{1}=3\) ( x ≥ 0 ; x # 1 )

⇒ \(A_{Max}=3."="\) ⇔ \(x=0\left(TM\right)\)

P = (\(\dfrac{1}{\sqrt{x}-1}\) - \(\dfrac{1}{\sqrt{x}}\)) : (\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\) - \(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)) với  0 < \(x\) ≠ 1; 4

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

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

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

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

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

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

b, P = \(\dfrac{1}{4}\)

⇒ \(\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{3x}\)  = \(\dfrac{1}{4}\)

⇒4\(x\) - 8\(\sqrt{x}\) = 3\(x\)

⇒ 4\(x\) - 8\(\sqrt{x}\) - 3\(x\) = 0

     \(x\) - 8\(\sqrt{x}\)   = 0

      \(\sqrt{x}\).(\(\sqrt{x}\) - 8) = 0

       \(\left[{}\begin{matrix}x=0\\\sqrt{x}=8\end{matrix}\right.\)

      \(\left[{}\begin{matrix}x=0\\x=64\end{matrix}\right.\)

      \(x=0\) (loại)

      \(x\) = 64

Lời giải:

a. \(P=\frac{\sqrt{x}-(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)}: \frac{(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}-1)}\)

\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}: \frac{x-1-(x-4)}{(\sqrt{x}-2)(\sqrt{x}-1)}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{3}{(\sqrt{x}-1)(\sqrt{x}-2)}\\ =\frac{1}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-1)(\sqrt{x}-2)}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)

b.

\(P=\frac{\sqrt{x}-2}{3\sqrt{x}}=\frac{1}{4}\\ \Rightarrow 4(\sqrt{x}-2)=3\sqrt{x}\\ \Leftrightarrow \sqrt{x}=8\Leftrightarrow x=64\) 

(thỏa mãn) 

c.

Tại $x=4+2\sqrt{3}=(\sqrt{3}+1)^2\Rightarrow \sqrt{x}=\sqrt{3}+1$
Khi đó:

$P=\frac{\sqrt{3}+1-2}{3(\sqrt{3}+1)}=\frac{2-\sqrt{3}}{3}$

`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

\(A=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\)  \(ĐKXĐ:x\ge0;x\ne1;x\ne4\)

\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\sqrt{x}+1}\right]:\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}-4}{x-1}\right]\)

\(A=\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\left[\frac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)

\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}:\frac{x-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

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

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

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

b)theo bài ra: \(A=\frac{1}{\sqrt{x}}\)

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

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

\(\Leftrightarrow x-\sqrt{x}-\sqrt{x}-2=0\)

\(\Leftrightarrow x-2\sqrt{x}-2=0\)

\(\Leftrightarrow x-2\sqrt{x}+1-3=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2-\left(\sqrt{3}\right)^2=0\)

\(\Leftrightarrow\left(\sqrt{x}-1-\sqrt{3}\right)\left(\sqrt{x}-1+\sqrt{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1-\sqrt{3}=0\\\sqrt{x}-1+\sqrt{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=\sqrt{3}+1\\\sqrt{x}=1-\sqrt{3}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\left(\sqrt{3}+1\right)^2\\x=\left(1-\sqrt{3}\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3+2\sqrt{3}+1\\x=3-2\sqrt{3}+1\end{cases}}\)

vậy......