Bài I

\(1,\\ a,A=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}-\dfrac{2\left(\sqrt{3}+1\right)}{2}=\sqrt{3}+2-\sqrt{3}-1=1\\ b,B=\left(\tan20^0.\tan70^0\right).\left(\tan30^0.\tan60^0\right).\left(\tan40^0.\tan50^0\right)\\ B=\left(\tan20^0.\cot20^0\right).\left(\tan30^0.\cot30^0\right).\left(\tan40^0.\cot40^0\right)=1.1.1=1\\ 2,\\ ĐK:x\ge0\\ PT\Leftrightarrow\left|\sqrt{x}-2\right|=3\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=3\\2-\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=25\left(tm\right)\\\sqrt{x}=-1\left(ktm\right)\end{matrix}\right.\)

Bài II:

\(1,A=\dfrac{3\cdot3-4}{3+1}=\dfrac{5}{4}\\ 2,B=\dfrac{2x+\sqrt{x}-4-x+4+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ 3,P=AB=\dfrac{3\sqrt{x}-4}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{3\sqrt{x}-4}{\sqrt{x}-1}\\ P\ge2\Leftrightarrow\dfrac{3\sqrt{x}-4}{\sqrt{x}-1}-2\ge0\\ \Leftrightarrow\dfrac{3\sqrt{x}-4-2\sqrt{x}+2}{\sqrt{x}-1}\ge0\\ \Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\ge0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}\ge2\\\sqrt{x}< 1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>4\\0\le x< 1\end{matrix}\right.\)

\(a,\Leftrightarrow x^2-2x-x^2+1=0\\ \Leftrightarrow-2x+1=0\Leftrightarrow x=\dfrac{1}{2}\\ b,\Leftrightarrow\left(2x-1-x-4\right)\left(2x-1+x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(3x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

\(0\le36x^2\le9\Leftrightarrow-3\le6x\le3\)

Áp dụng tc dtsbn:

\(2\widehat{A}=3\widehat{B};\dfrac{\widehat{B}}{1}=\dfrac{\widehat{C}}{2}\Rightarrow\dfrac{\widehat{A}}{3}=\dfrac{\widehat{B}}{2};\dfrac{\widehat{B}}{1}=\dfrac{\widehat{C}}{2}\\ \Rightarrow\dfrac{\widehat{A}}{3}=\dfrac{\widehat{B}}{2}=\dfrac{\widehat{C}}{4}=\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{3+2+4}=\dfrac{180^0}{9}=20^0\\ \Rightarrow\left\{{}\begin{matrix}\widehat{A}=60^0\\\widehat{B}=40^0\\\widehat{C}=80^0\end{matrix}\right.\)

\(a,ĐK:x\ne0;x\ne5\\ B=\dfrac{x^2-25+2x^2-12x-x^2+8x+25}{2x\left(x-5\right)}=\dfrac{2x\left(x-2\right)}{2x\left(x-5\right)}=\dfrac{x-2}{x-5}\\ b,x=3\Leftrightarrow A=\dfrac{3+6}{5-3}=\dfrac{9}{2}\\ c,\text{Câu a}\\ d,E=B-A=\dfrac{x-2}{x-5}+\dfrac{x+6}{x-5}=\dfrac{2x+4}{x-5}=\dfrac{2\left(x-5\right)+14}{x-5}=2+\dfrac{14}{x-5}\in Z\\ \Leftrightarrow x-5\inƯ\left(14\right)=\left\{-14;-7;-2;-1;1;2;7;14\right\}\\ \Leftrightarrow x\in\left\{-9;-2;3;4;6;7;12;19\right\}\)

vế trái vế phải

\(VT=\left|3x+4\right|+\left|1-3x\right|\ge\left|3x+4+1-3x\right|=5\\ VP\le\dfrac{20}{3\cdot0+4}=5\\ \Rightarrow VT\ge5\ge VP\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}\left(3x+4\right)\left(1-3x\right)\ge0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{4}{3}\le x\le\dfrac{1}{3}\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\)

\(=\dfrac{x+12+x-2}{x-2}\cdot\dfrac{x\left(x-2\right)}{\left(5-x\right)\left(5+x\right)}=\dfrac{2x\left(x+5\right)}{\left(5-x\right)\left(5+x\right)}=\dfrac{2x}{5-x}\)