b: \(A=\dfrac{P}{Q}=\dfrac{x+3}{\sqrt{x}-2}:\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{x+3}{\sqrt{x}}=\sqrt{x}+\dfrac{3}{\sqrt{x}}\ge2\sqrt{3}\)

Dấu '=' xảy ra khi \(x=3\)

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

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

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

b: Để A là số nguyên thì \(\left(\sqrt{x}-1\right)^2⋮\left(\sqrt{x}+1\right)\left(-x+\sqrt{x}+1\right)\)

=>x=0

a: 

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

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

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

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

b: Để A là số nguyên thì \(\sqrt{x}\left(-\sqrt{x}+3\right)⋮x-\sqrt{x}-1\)

=>\(-x+3\sqrt{x}⋮x-\sqrt{x}-1\)

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

=>\(x=0\)

a: \(P=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)

b: \(P=\dfrac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\)

\(\Leftrightarrow P=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6>=2\cdot5-6=10-6=4\)

Dấu = xảy ra khi x=4

Sửa đề: Tìm giá trị lớn nhất

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)

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

\(\sqrt{x}-2>=-2\forall x\) thỏa mãn ĐKXĐ

=>\(\dfrac{2}{\sqrt{x}-2}< =-1\forall x\) thỏa mãn ĐKXĐ

=>\(\dfrac{2}{\sqrt{x}-2}+1< =-1+1=0\forall x\) thỏa mãn ĐKXĐ

=>\(C< =0\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x=0

Câu 3

a, ĐKXĐ: x>0, x\(\ne\)4

M=( \(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\)). \(\dfrac{\sqrt{x}+2}{\sqrt{4x}}\)

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

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

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

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

b, Thay x= \(6+4\sqrt{2}\) ( x>0, x\(\ne\)4) ta có:

M= \(\dfrac{\sqrt{6+4\sqrt{2}}}{\sqrt{6+4\sqrt{2}}-2}\)

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

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

Vậy khi x= \(6+4\sqrt{2}\) thì M= \(1+\sqrt{2}\)

c, Để M<1 <=> \(\dfrac{\sqrt{x}}{\sqrt{x}-2}< 1\)

<=> \(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-2}< 0\)

<=> \(\dfrac{2}{\sqrt{x}-2}< 0\)

Vì 2>0 <=> \(\sqrt{x}-2< 0\)

<=> \(\sqrt{x}< 2\)

<=> x<4

Vậy để M<1 thì 0<x<4

<=>

Câu 2

a, \(\sqrt{3x+2}=5\) (x\(\ge\dfrac{-2}{3}\))

<=> \(\sqrt{3x+2}=\sqrt{25}\)

<=> 3x+2=25

<=> 3x= 23

<=> x=\(\dfrac{23}{3}\)

Vậy S= \(\left\{\dfrac{23}{3}\right\}\)

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

\(dk:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

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

a)

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

b) tồn tại \(\sqrt{P}\Rightarrow\dfrac{x}{\sqrt{x}-1}\ge0\) \(\Leftrightarrow x>1\)

\(\left\{{}\begin{matrix}x>1\\P=\dfrac{x}{\sqrt{x}-1}=\left(\sqrt{x}-1\right)+\dfrac{1}{\sqrt{x}-1}+2\ge2+2=4\end{matrix}\right.\)đẳng thức khi x =\(\left(\sqrt{x}-1\right)^2=1\Rightarrow x=4\) thỏa mãn

GTNN \(\sqrt{P}=2\)