ĐKXĐ: x>=0; x<>1/4

Ta có: \(A=\frac{\sqrt{x}+1}{2\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{x+6\sqrt{x}+2}{2x+5\sqrt{x}-3}\)

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

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+\sqrt{x}\left(2\sqrt{x}-1\right)-x-6\sqrt{x}-2}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+4\sqrt{x}+3+2x-\sqrt{x}-x-6\sqrt{x}-2}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{2x-3\sqrt{x}+1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

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

Ta có: P=A*B

\(=\frac{\sqrt{x}-1}{\sqrt{x}+3}\cdot\frac{\sqrt{x}+3}{x+8}=\frac{\sqrt{x}-1}{x+8}\)

=>\(\frac{1}{P}=\frac{x+8}{\sqrt{x}-1}=\frac{x-1+9}{\sqrt{x}-1}=\sqrt{x}+1+\frac{9}{\sqrt{x}-1}=\sqrt{x}-1+\frac{9}{\sqrt{x}-1}+2\ge2\cdot\sqrt{\left(\sqrt{x}-1\right)\cdot\frac{9}{\sqrt{x}-1}}+2=2\cdot3+2=8\forall x\) thỏa mãn ĐKXĐ

=>\(P\le\frac18\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\left(\sqrt{x}-1\right)^2=9;\sqrt{x}-1>0\)

=>\(\sqrt{x}-1=3\)

=>\(\sqrt{x}=4\)

=>x=16(nhận)

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

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

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

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

b. Đặt \(B=A-2x\)

\(B=\sqrt{x}-1-2x=-2\left(\sqrt{x}-\dfrac{1}{4}\right)^2-\dfrac{7}{8}\le-\dfrac{7}{8}\)

\(B_{max}=-\dfrac{7}{8}\) khi \(\sqrt{x}-\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{1}{16}\)

Áp dụng BĐT cosi:

\(A=\sqrt{\left(2x+1\right)\left(x+2\right)}+2\sqrt{x+3}-2x\\ A\le\dfrac{2x+1+x+2}{2}+\dfrac{4+x+3}{2}-2x\\ A\le\dfrac{3x+3}{2}+\dfrac{x+7}{2}-2x=\dfrac{3x+3+x+7-4x}{2}=5\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}2x+1=x+2\\4=x+3\end{matrix}\right.\Leftrightarrow x=1\)

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

\(P\le\dfrac{1}{2}\left(x+2+2x+1\right)+\dfrac{1}{2}\left(4+x+3\right)-2x=5\)

\(P_{max}=5\) khi \(x=1\)

Lời giải:

ĐK: \(x> 0; x\ne 1; x\ne \frac{1}{4}\)

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

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

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

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

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

b) ĐK chưa để tìm GTLN, GTNN

c) Tại \(x=7+2\sqrt{10}\Rightarrow P=\frac{6+2\sqrt{10}}{7+2\sqrt{10}}\)

d) \(P=\frac{x-1}{x}=1-\frac{1}{x}< 1\) với moi \(x>0\) nên không tồn tại giá trị của $x$ để $P>1$

1:

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

căn x+1>=1

=>2/căn x+1<=2

=>-2/căn x+1>=-2

=>A>=-2+1=-1

Dấu = xảy ra khi x=0

b: 

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

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