rút gọn R ?

a: ta có: \(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

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

\(=\dfrac{4\sqrt{x}}{\sqrt{x}-3}\)

a) \(P=\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}=\dfrac{4x}{\sqrt{x}-3}\)

\(\left(x\ge0;x\ne4;9\right)\)

b)\(P=-1\Leftrightarrow4x+\sqrt{x}-3=0\Leftrightarrow\sqrt{x}=\dfrac{3}{4}\Leftrightarrow x=\dfrac{9}{16}\)

c) bpt đưa về dạng \(4mx>x+1\Leftrightarrow\left(4x-1\right)x>1\)

Nếu \(4m-1\le0\) thì tập nghiệm không thể chứa mọi giá trị \(x>9\); Nếu \(4m-1>0\) thì tập nghiệm bpt là \(x>\dfrac{1}{4m-1}\). Do đó bpt tm mọi \(x>9\Leftrightarrow9\ge\dfrac{1}{4m-1}\) và \(4m-1>0\). ta có \(m\ge\dfrac{5}{18}\)

a: \(Q=\left(\dfrac{4\sqrt{x}}{x+2\sqrt{x}}+\dfrac{8\sqrt{x}}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\left(\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\left(\dfrac{4}{\sqrt{x}+2}-\dfrac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)

\(=\dfrac{4\left(\sqrt{x}-2\right)-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\)

\(=\dfrac{-4\sqrt{x}-8}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{-\left(\sqrt{x}-3\right)}=\dfrac{4\sqrt{x}}{\sqrt{x}-3}\)

b: Q<4

=>Q-4<0

=>\(\dfrac{4\sqrt{x}}{\sqrt{x}-3}-4< 0\)

=>\(\dfrac{4\sqrt{x}-4\sqrt{x}+12}{\sqrt{x}-3}< 0\)

=>\(\dfrac{12}{\sqrt{x}-3}< 0\)

=>\(\sqrt{x}-3< 0\)

=>0<=x<9

Kết hợp ĐKXĐ, ta được: 0<x<9 và x<>4

\(a,Q=\left(\dfrac{4\sqrt{x}}{x+2\sqrt{x}}+\dfrac{8\sqrt{x}}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\\ =\left(\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2}{\sqrt{x}}\right)\\ =\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{4x-8\sqrt{x}-8x}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)

\(=\dfrac{-4x-8\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\\ =\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\\ =\dfrac{-4\sqrt{x}}{3-\sqrt{x}}\)

`b,` Để `Q<4` ta có :

\(\dfrac{-4\sqrt{x}}{3-\sqrt{x}}< 4\\ \Leftrightarrow\dfrac{-4\sqrt{x}}{3-\sqrt{x}}-4< 0\\ \Leftrightarrow\dfrac{-4\sqrt{x}-4\left(3-\sqrt{x}\right)}{3-\sqrt{x}}< 0\\ \Leftrightarrow-4\sqrt{x}-12+4\sqrt{x}< 0\\ \Leftrightarrow-12< 0\left(luon.dung\right)\)

c??

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

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

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

Để P nguyên thì \(\sqrt{x}-2\in\left\{-1;1\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{1;3\right\}\)

hay \(x\in\left\{1;9\right\}\)

Mik ko hiểu đề

a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

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

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

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

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

\(=\sqrt{x}\)

b) Để P>4 thì \(\sqrt{x}>4\)

hay x>16

Kết hợp ĐKXĐ, ta được: x>16

Vậy: Khi x>16 thì P>4

a: \(=\dfrac{4x-8\sqrt{x}+8x}{x-4}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

b: \(m\left(\sqrt{x}-3\right)\cdot B>x+1\)

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

=>\(-12m\cdot x\sqrt{x}+8xm>x\sqrt{x}+2x+\sqrt{x}+2\)

=>\(x\sqrt{x}\left(-12m-1\right)+x\left(8m-2\right)-\sqrt{x}-2>0\)

Để BPT luôn đúng thì m<-0,3

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

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

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

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

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

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

Để A>1 thì A-1>0

\(\Leftrightarrow\dfrac{4\sqrt{x}-3-\sqrt{x}+2}{\sqrt{x}-2}>0\)

\(\Leftrightarrow\dfrac{3\sqrt{x}-1}{\sqrt{x}-2}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}3\sqrt{x}-1\le0\\\sqrt{x}-2>0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< x\le\dfrac{1}{9}\\x>4\end{matrix}\right.\)

Đề hơi sai sai ý bạn ơi

a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

b) Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)

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

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

d) Để A>0 thì \(\sqrt{x}-2>0\)

hay x>4

a) đk: \(x\ne0;4\); \(x>0\)

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

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

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

b) Để P < \(\dfrac{1}{2}\)

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

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

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

<=> \(\sqrt{x}< 2\)

<=> x < 4

<=> 0 < x < 4

c, Tìm gt nguyên của x để P có gt nguyên

a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)