\(sin3x+sin\left(5x-\dfrac{\pi}{6}\right)=0.\\ TXD:D=R.\\ \Leftrightarrow sin3x=-sin\left(5x-\dfrac{\pi}{6}\right).\\ \Leftrightarrow sin3x=sin\left(\dfrac{\pi}{6}+5x\right).\\ \Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{6}+5x+k2\pi.\\3x=\pi-\dfrac{\pi}{6}-5x+k2\pi.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=\dfrac{\pi}{6}+k2\pi\\8x=\dfrac{5}{6}\pi+k2\pi.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{12}\pi-k\pi.\\x=\dfrac{5}{48}\pi+\dfrac{k\pi}{4}.\end{matrix}\right.\)

\(\dfrac{x}{y}=\dfrac{3}{7}.\\ \Rightarrow x=\dfrac{3}{7}y.\\ x-y=16.\\\Rightarrow\dfrac{3}{7}y-y=16.\\ \Rightarrow y=-28.\\ \Rightarrow x=-12.\)

\(\dfrac{x}{1,8}=\dfrac{y}{3,2}.\\ \Rightarrow\dfrac{x}{y}=\dfrac{1,8}{3,2}=\dfrac{9}{16}.\\ \Rightarrow x=\dfrac{9}{16}y.\\ y-x=7.\\ \Rightarrow y-\dfrac{9}{16}y=7.\\ \Leftrightarrow y=16.\\ \Leftrightarrow x=9.\)

\(\dfrac{x}{5}=\dfrac{y}{8}.\\ \Rightarrow\dfrac{x}{y}=\dfrac{5}{8}.\\ \Rightarrow x=\dfrac{5}{8}y.\\ x+2y=42.\\ \Rightarrow\dfrac{5}{8}y+2y=42.\\ \Leftrightarrow y=16.\\ \Rightarrow x=10.\)

\(\dfrac{x}{5}=\dfrac{y}{7}.\\ \Rightarrow\dfrac{x}{y}=\dfrac{5}{7}.\\ \Rightarrow x=\dfrac{5}{7}y.\\ x.y=35.\\ \Rightarrow\dfrac{5}{7}y.y=35.\\ \Leftrightarrow y^2=49.\\ \Leftrightarrow u=\pm7.\\ \Rightarrow x=\pm5.\)

Danh từ: Sân khấu; vòm trời; lá; cành; nhạc công; kèn.

Động từ: thổi.

Tính từ: biếc; náo nhiệt.

\(sin\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)=\dfrac{-1}{2}.\\ TXD:D=R.\\ \Leftrightarrow sin\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)=sin\dfrac{-\pi}{6}.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}-\dfrac{\pi}{4}=\dfrac{-\pi}{6}+k2\pi.\\\dfrac{x}{2}-\dfrac{\pi}{4}=\pi-\dfrac{-\pi}{6}+k2\pi.\end{matrix}\right.\) \(\left(k\in Z\right).\)

\(\Leftrightarrow\left[{}\begin{matrix}6x-3\pi=-2\pi+k24\pi.\\6x-3\pi=12\pi-2\pi+k24\pi.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k4\pi.\\x=\dfrac{13}{6}\pi+k4\pi.\end{matrix}\right.\)

\(\dfrac{x}{7}=\dfrac{y}{3}.\\ \Rightarrow3x=7y.\\ \Rightarrow x=\dfrac{7}{3}y.\\ y-x=2.\\ \Rightarrow y-\dfrac{7}{3}y=2.\\ \Leftrightarrow-\dfrac{4}{3}y=2.\\ \Leftrightarrow y=\dfrac{-3}{2}.\\ \Rightarrow x=\dfrac{-7}{2}.\)

+) Đường cao AH:

\(\left\{{}\begin{matrix}\overrightarrow{n_{AH}}=\overrightarrow{BC}=\left(1;-1\right).\\quaB=\left(0;4\right).\end{matrix}\right.\)

\(\Rightarrow PTTQ:1\left(x-0\right)-1\left(y-4\right)=0.\\ \Leftrightarrow x-y+4=0.\)

+) Đường cao BH:

\(\left\{{}\begin{matrix}\overrightarrow{n_{BH}}=\overrightarrow{CA}=\left(1;-3\right).\\quaA=\left(2;0\right).\end{matrix}\right.\)

\(\Rightarrow PTTQ:1\left(x-2\right)-3\left(y-0\right)=0.\\ \Leftrightarrow x-3y-2=0.\)

+) Tọa độ trực tâm H của tam giác ABC là giao điểm của 2 đường cao AH và BH nên ta có hệ phương trình sau:

\(\left\{{}\begin{matrix}x-y+4=0.\\x-3y-2=0.\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=-4.\\x-3y=2.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-7.\\y=-3.\end{matrix}\right.\)

\(\Rightarrow\) Tọa độ trực tâm H của tam giác ABC là \(H\left(-7;-3\right).\)

\(B7:\\ a,\dfrac{5}{21}+0,5-\dfrac{19}{23}+\dfrac{16}{21}-\dfrac{4}{23}.\\ =\left(\dfrac{5}{21}+\dfrac{16}{21}\right)-\left(\dfrac{19}{23}+\dfrac{4}{23}\right)+0,5.\\ =1-1+0,5=0,5.\)

\(B8:\\ a,\left(\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{17}{4}-\dfrac{3}{4}.\\ =\dfrac{17}{12}:\dfrac{17}{4}-\dfrac{3}{4}.\\ =\dfrac{1}{3}-\dfrac{3}{4}=-\dfrac{5}{12}.\\ b,\left(-5\right)^2.\dfrac{7}{45}+\left(-5\right)^2.\dfrac{11}{45}.\\ =\left(-5\right)^2\left(\dfrac{7}{45}+\dfrac{11}{45}\right).\\ =25.\dfrac{2}{5}=10.\)

Xét \(\Delta ACD:\)

+ CH là trung tuyến (Do H là trung điểm của AD).

+ CH là đường cao \(\left(CH\perp BC\right).\)

\(\Rightarrow\) \(\Delta ACD\) cân tại C.