a, ĐKXĐ: \(x\ge0,\)

b, ĐKXĐ: \(x\ge0,x\ne1\)

c, ĐKXĐ: \(x\ge0,x\ne4\)

d,ĐKXĐ:\(x\ge0,x\ne9,x\ne4\)

e,ĐKXĐ:\(x\ge0,x\ne1,x\ne4\)

\(a,B=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\left(x>0;x\ne6\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x+3\sqrt{x}+\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\\)

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

`b,` Tớ tính mãi ko ra, xl cậu nha=')

Ta có: \(P=A\cdot B\) (ĐK: \(x>0;x\ne4\))

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

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

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

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

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

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+9}\)

Với x > 0; x ≠ 4 thì \(\sqrt{P}< \dfrac{1}{3}\Leftrightarrow P< \dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+9}< \dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+9}-\dfrac{1}{9}< 0\)

\(\Leftrightarrow\dfrac{9\left(\sqrt{x}-1\right)}{9\left(\sqrt{x}+9\right)}-\dfrac{\sqrt{x}+9}{9\left(\sqrt{x}+9\right)}< 0\)

\(\Leftrightarrow\dfrac{9\sqrt{x}-9-\sqrt{x}-9}{9\sqrt{x}+81}< 0\)

\(\Leftrightarrow\dfrac{8\sqrt{x}-18}{9\sqrt{x}+18}< 0\)

Ta thấy: \(9\sqrt{x}+18>0\forall x\)

\(\Rightarrow8\sqrt{x}-18< 0\)

\(\Rightarrow\sqrt{x}< \dfrac{18}{8}\)

\(\Rightarrow\sqrt{x}< \dfrac{9}{4}\Leftrightarrow x< \dfrac{81}{16}\)

Kết hợp với điều kiện, ta được: \(0< x\le5\)\(;x\ne4\)

\(\Rightarrow x\in\left\{1;2;3;5\right\};x\in Z\) thì \(\sqrt{P}< \dfrac{1}{3}\)

#Urushi

Ta có: \(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-3x+8\sqrt{x}-5-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

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

\(\Leftrightarrow A-\dfrac{2}{3}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{2}{3}\)

\(\Leftrightarrow A-\dfrac{2}{3}=\dfrac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow A-\dfrac{2}{3}=\dfrac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\le0\)

\(\Leftrightarrow A\le\dfrac{2}{3}\)

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

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

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

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

Lời giải:
Xét số hạng tổng quát: 

\(\frac{\sqrt{n+1}-\sqrt{n}}{n+(n+1)}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n(n+1)}}=\frac{1}{2}(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})\) theo BĐT Cô-si.

Do đó:
\(x< \frac{1}{2}\left[\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\right]=\frac{1}{2}(1-\frac{1}{\sqrt{100}})< \frac{1}{2}\)

Ta có đpcm.

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

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

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

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

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

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

c) Để \(P< -\dfrac{1}{2}\) thì \(P+\dfrac{1}{2}< 0\)

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

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

\(\Leftrightarrow\sqrt{x}-3< 0\)

\(\Leftrightarrow x< 9\)

Kết hợp ĐKXĐ, ta được: \(0\le x< 9\)