\(\dfrac{1}{x^2+x}=\dfrac{x-1}{x\left(x-1\right)\left(x+1\right)};\dfrac{x^2-4}{x^2-1}=\dfrac{x\left(x^2-4\right)}{x\left(x-1\right)\left(x+1\right)}\\ \dfrac{1}{y-1}-\dfrac{1}{y}=\dfrac{y-y+1}{y\left(y-1\right)}=\dfrac{1}{y\left(y-1\right)}\)

Ta có hệ số góc là 3>0 nên tạo với Ox 1 góc nhọn

\(\Rightarrow\tan\alpha=3\)

\(1m\approx3,28ft\\ \Rightarrow1ft\approx\dfrac{1}{3,28}\approx0,305m\\ \Rightarrow2ft\approx2\cdot0,305=0,61\left(m\right)\)

\(\dfrac{x+3}{x^2-3x}+\dfrac{3}{x^2+3x}+\dfrac{2x-18}{x^2-9}\\ =\dfrac{x^2+6x+9+3x-9+2x^2-18x}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3x^2-9x}{x\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

PT giao Ox: \(y=0\Leftrightarrow x=2\Leftrightarrow A\left(2;0\right)\Leftrightarrow OA=2\)

PT giao Oy: \(x=0\Leftrightarrow y=-4\Leftrightarrow B\left(0;-4\right)\Leftrightarrow OB=4\)

\(\Leftrightarrow S_{AIB}=\dfrac{1}{2}S_{AOB}=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot OA\cdot OB=\dfrac{1}{4}\cdot2\cdot4=2\left(đvdt\right)\)

\(AC=\tan B\cdot AB=\dfrac{5}{12}\cdot6=\dfrac{5}{2}=2,5\left(cm\right)\\ \Rightarrow BC=\sqrt{AC^2+AB^2}=6,5\left(cm\right)\left(pytago\right)\)

Áp dụng PTG, ta có: \(BD=\sqrt{BC^2+CD^2}=10\left(cm\right)\)

Áp dụng HTL, ta có: \(AH=\dfrac{BC\cdot CD}{BD}=4,8\left(cm\right)\)

\(\dfrac{2x+2}{x+3}=\dfrac{2\left(x+3\right)-4}{x+3}=2-\dfrac{4}{x+3}\in Z\\ \Leftrightarrow x+3\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\\ \Leftrightarrow x\in\left\{-7;-5;-4;-2;-1;1\right\}\)

\(\Rightarrow\dfrac{3x}{60}=\dfrac{4y}{60}=\dfrac{5z}{60}\Rightarrow\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x+y+z}{20+15+12}=\dfrac{141}{47}=3\\ \Rightarrow\left\{{}\begin{matrix}x=60\\y=45\\z=36\end{matrix}\right.\)