1: Khi x=36 thì \(A=\dfrac{6}{2\cdot6-4}=\dfrac{6}{12-4}=\dfrac{6}{8}=\dfrac{3}{4}\)

2: 

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

\(C=B:A\)

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

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

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

3: \(C\cdot\sqrt{x}< \dfrac{4}{3}\)

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

=>\(\dfrac{2\sqrt{x}\cdot3-4\left(\sqrt{x}+2\right)}{3\left(\sqrt{x}+2\right)}< 0\)

=>\(6\sqrt{x}-4\sqrt{x}-8< 0\)

=>\(2\sqrt{x}-8< 0\)

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

=>\(0< =x< 16\)

Kết hợp ĐKXĐ của C, ta được: \(\left\{{}\begin{matrix}0< x< 16\\x< >4\end{matrix}\right.\)

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

Khi x=9 thì \(B=\dfrac{\sqrt{9}+1}{\sqrt{9}+2}\)

\(=\dfrac{3+1}{3+2}=\dfrac{4}{5}\)

b: \(A=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{6+\sqrt{x}}{x-4}\)

\(=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+6}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}\)

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

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

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

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

c: P=A/B

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

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

\(=\dfrac{-2}{\sqrt{x}+1}< 0\)

=>P<2

\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{x-9}\right)\cdot\dfrac{2\sqrt{x}+6}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}-3+2\sqrt{x}}{x-9}\cdot\dfrac{2\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)

\(=\dfrac{3\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\cdot\dfrac{2}{\sqrt{x}-3}=\dfrac{6}{\sqrt{x}-3}\)

a) Ta có: \(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)

\(=\dfrac{6\left(\sqrt{5}-1\right)}{4}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{4}}-2\sqrt{5}\)

\(=\dfrac{3}{2}\left(\sqrt{5}-1\right)+\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-2\sqrt{5}\)

\(=\dfrac{3}{2}\sqrt{5}-\dfrac{3}{2}-2\sqrt{5}+\dfrac{\sqrt{5}+1}{2}\)

\(=-\dfrac{1}{2}\sqrt{5}-\dfrac{3}{2}+\dfrac{1}{2}\sqrt{5}+\dfrac{1}{2}\)

=-1

Bài 1: 

a) Thay \(x=\dfrac{1}{4}\)vào B, ta được:

\(B=1:\left(\dfrac{1}{4}\cdot\dfrac{1}{2}+27\right)=1:\left(27+\dfrac{1}{8}\right)=\dfrac{8}{217}\)

b) Ta có: \(A=\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(=\dfrac{x-9+\sqrt{x}+3-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

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

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

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

c) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)

\(\Leftrightarrow\dfrac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)

\(\Leftrightarrow3-\sqrt{x}>0\)

\(\Leftrightarrow\sqrt{x}< 3\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne4\end{matrix}\right.\)

a: Thay \(x=\dfrac{1}{4}\) vào A, ta được:

\(A=\left(\dfrac{1}{2}+1\right):\left(\dfrac{1}{2}-2\right)=\dfrac{3}{2}:\dfrac{-3}{2}=-1\)

b: Ta có: \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\)

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

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

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

c: Để B là số tự nhiên thì \(\sqrt{x}+4⋮\sqrt{x}-2\)

\(\Leftrightarrow\sqrt{x}-2\in\left\{1;2;3;6\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{3;4;5;8\right\}\)

hay \(x\in\left\{16;25;64\right\}\)

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé.

a: \(A=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)

\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=5-1=4\)

b: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)

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

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

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

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

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

c: Khi x=9 thì \(B=\dfrac{-2}{\sqrt{9}+1}=\dfrac{-2}{3+1}=-\dfrac{2}{4}=-\dfrac{1}{2}\)

d: |B|=A

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

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

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

=>\(\sqrt{x}=-\dfrac{1}{2}\)(loại) hoặc \(\sqrt{x}=-\dfrac{3}{2}\)(loại)

Bạn xem lại xem đã biết biểu thức đúng chưa vậy?

\(P=A:B=\dfrac{1-\sqrt{x}}{\sqrt{x}-2}:\dfrac{2\sqrt{x}}{\sqrt{x}-2}=\dfrac{1-\sqrt{x}}{2\sqrt{x}}\)

Có: \(\left|P+1\right|< 3P\left(ĐK:x>0\right)\)

\(\Leftrightarrow\left|\dfrac{1-\sqrt{x}}{2\sqrt{x}}+1\right|< 3.\dfrac{1-\sqrt{x}}{2\sqrt{x}}\\ \Leftrightarrow\left|\dfrac{1-\sqrt{x}+2\sqrt{x}}{2\sqrt{x}}\right|< \dfrac{3-3\sqrt{x}}{2\sqrt{x}}\\ \Leftrightarrow\left|\dfrac{\sqrt{x}+1}{2\sqrt{x}}\right|< \dfrac{3-3\sqrt{x}}{2\sqrt{x}}\)

Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\) nên:

\(\left|\dfrac{\sqrt{x}+1}{2\sqrt{x}}\right|< \dfrac{3-3\sqrt{x}}{2\sqrt{x}}\\ \Leftrightarrow\dfrac{\sqrt{x}+1-3+3\sqrt{x}}{2\sqrt{x}}< 0\\ \Leftrightarrow\dfrac{4\sqrt{x}-2}{2\sqrt{x}}< 0\\ \Leftrightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}}< 0\\ \Rightarrow\left\{{}\begin{matrix}\sqrt{x}>0\\2\sqrt{x}-1< 0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x>0\\x< \dfrac{1}{4}\end{matrix}\right.\)

\(\Rightarrow0< x< \dfrac{1}{4}\)