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

<=> \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

<=> \(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

b. Khi \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\) => \(\sqrt{x}=2+\sqrt{3}\)

=> \(P=\dfrac{2+\sqrt{3}+2}{7+4\sqrt{3}-2\left(2+\sqrt{3}\right)}=\dfrac{4+\sqrt{3}}{7+4\sqrt{3}-4-2\sqrt{3}}=\dfrac{4+\sqrt{3}}{3+2\sqrt{3}}=\dfrac{5\sqrt{3}-6}{3}\)

check giùm mik

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

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

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

b: Thay \(x=7+4\sqrt{3}\) vào P, ta được:

\(P=\dfrac{2+\sqrt{3}+2}{\sqrt{3}\left(2+\sqrt{3}\right)}=\dfrac{-6+5\sqrt{3}}{3}\)

Sửa đề: \(B=\dfrac{x-3\sqrt{x}+4}{x-2}-\dfrac{1}{\sqrt{x}-2}\)

Ta có: \(B=\dfrac{x-3\sqrt{x}+4}{x-2}-\dfrac{1}{\sqrt{x}-2}\)

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

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

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

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

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

       \(:\left(1-\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\right)\)

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

\(:\left(\dfrac{\text{​​}\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\right)\)

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

\(.\left(\dfrac{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}\right)\)

\(A=1\)

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

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

\(\Leftrightarrow-\sqrt{x}-1=\sqrt{x}-1\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)

c) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\)

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

Kết hợp đk:

\(\Rightarrow x\in\left\{0\right\}\)

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

\(a,P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,P=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\\ \Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\\ c,P\in Z\Leftrightarrow\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}\in Z\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\in Z\\ \Leftrightarrow2⋮\sqrt{x}+1\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}+1\in\left\{1;2\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}\in\left\{0;1\right\}\\ \Leftrightarrow x\in\left\{0;1\right\}\)

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

Có \(\dfrac{2}{\sqrt{x}+1}>0\left(2>0;\sqrt{x}+1>0\right)\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}< 1\Leftrightarrow P< 1\)

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

Có \(\sqrt{x}+1\ge1\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\le2\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\ge1-2=-1\)

\(P_{min}=-1\Leftrightarrow x=0\)

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

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

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

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

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

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

\(Q=\left(\dfrac{1}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}-1}\right):\dfrac{1}{1-4x}\left(dkxd:x\ge0;x\ne\dfrac{1}{4}\right)\)

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

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

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

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

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

ĐK:\(x\ge0;x\ne9\)

a) \(P=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+x-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

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

b)\(P=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}=1+\dfrac{2}{\sqrt{x}+2}\le1+\dfrac{2}{0+2}=2\)

Dấu "=" xảy ra khi \(x=0\)

Vậy \(P_{max}=2\)

Đề hơi sai sai ý bạn ơi

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

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

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

\(\dfrac{2\sqrt{x}}{\sqrt{x}-2}.\dfrac{\sqrt{x}+2}{\sqrt{x}-2}=\dfrac{2\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)^2}\) (\(đk:x\ge0;x\ne\sqrt{2}\))

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

\(\)