4b.

\(\dfrac{\pi}{2}< a< \pi\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\dfrac{4}{5}\)

\(\Rightarrow tana=\dfrac{sina}{cosa}=-\dfrac{3}{4}\)

\(tan\left(a+\dfrac{\pi}{3}\right)=\dfrac{tana+tan\left(\dfrac{\pi}{3}\right)}{1-tana.tan\left(\dfrac{\pi}{3}\right)}=\dfrac{-\dfrac{3}{4}+\sqrt{3}}{1-\left(-\dfrac{3}{4}\right).\sqrt{3}}=...\)

c.

\(\dfrac{3\pi}{2}< a< 2\pi\Rightarrow cosa>0\Rightarrow cosa=\sqrt{1-sin^2a}=\dfrac{5}{13}\)

\(cos\left(\dfrac{\pi}{3}-a\right)=cos\left(\dfrac{\pi}{3}\right).cosa+sin\left(\dfrac{\pi}{3}\right).sina=\dfrac{1}{2}.\dfrac{5}{13}+\left(-\dfrac{12}{13}\right).\dfrac{\sqrt{3}}{2}=...\)

Lời giải:

Mình sẽ xác định giúp bạn tập hợp. Việc biểu diễn nó trên trục số thì có lẽ đơn giản rồi.

1. $(2;9)\cap (1;4)=(2;4)$

2. $(-\infty; -2)\cup (-3;5)=(-\infty; 5)$

3. $[3;1]\cap (-2;0)=$ \(\varnothing\)

4.\((-2;1)\setminus [0;3)=(-2;0)\)

5. \((-\infty; -2]\setminus (-3;5)=(-\infty; -3]\)

6. $[3;+\infty)\cap (-2;-0)=\varnothing$

7. $[1;2]\setminus [0;3)=[1;0)$

8.$[-4;3)\cup [3;5]=[-4;5]$

9. $(-\infty;3)\setminus (-4;-1)=\varnothing$

🤔

🌚

a.

D chia CB theo tỉ số \(k=2\Rightarrow\)\(\overrightarrow{DC}=2\overrightarrow{DB}\)

\(\Rightarrow\overrightarrow{DC}-\overrightarrow{DB}=\overrightarrow{DB}\Rightarrow\overrightarrow{DC}+\overrightarrow{BD}=\overrightarrow{DB}\)

\(\Rightarrow\overrightarrow{BC}=\overrightarrow{DB}\Rightarrow\overrightarrow{BC}+\overrightarrow{BD}=\overrightarrow{0}\)

\(\Rightarrow\) B là trung điểm CD hay D là điểm đối xứng C qua B

Do M là trung điểm AB \(\Rightarrow\overrightarrow{MA}=-\overrightarrow{MB}\)

\(\overrightarrow{CA}.\overrightarrow{CB}=\left(\overrightarrow{CM}+\overrightarrow{MA}\right)\left(\overrightarrow{CM}+\overrightarrow{MB}\right)=\left(2\overrightarrow{CI}-\overrightarrow{MB}\right)\left(2\overrightarrow{CI}+\overrightarrow{MB}\right)\)

\(=4\overrightarrow{CI}^2-\overrightarrow{MB}^2=4CI^2-MB^2\)

b.

\(2\left(1-cos^2C\right)+3cosC=0\Leftrightarrow-2cos^2C+3cosC+2=0\Rightarrow\left[{}\begin{matrix}cosC=2>1\left(loại\right)\\cosC=-\dfrac{1}{2}\end{matrix}\right.\)

Mặt khác: \(cosC=\dfrac{AC^2+BC^2-AB^2}{2AC.BC}=\dfrac{20-AB^2}{16}=-\dfrac{1}{2}\)

\(\Rightarrow AB^2=28\Rightarrow AB=2\sqrt{7}\)

\(\Rightarrow cosB=\dfrac{AB^2+BC^2-AC^2}{2AB.BC}=\dfrac{5\sqrt{7}}{14}\)

\(\Rightarrow CH=BC.\sqrt{1-cos^2B}=\dfrac{2\sqrt{21}}{7}\)

\(BM=\dfrac{1}{2}AB=\sqrt{7}\Rightarrow CM=\sqrt{BM^2+BC^2-2BM.BC.cosB}=\sqrt{3}\)

Áp dụng công thức trung tuyến:

\(BI=\dfrac{\sqrt{2\left(BM^2+BC^2\right)-CM^2}}{2}=...\)

3: \(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+2\left(x>=\dfrac{1}{2}\right)\\2x-1=-x-2\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-\dfrac{1}{3}\left(nhận\right)\end{matrix}\right.\)

\(1,ĐK:x\ge1\\ PT\Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\\ 2,\Leftrightarrow2x-5=x^2-8x+16\left(x\ge4\right)\\ \Leftrightarrow x^2-10x+21=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=7\left(tm\right)\end{matrix}\right.\\ 3,\Leftrightarrow\left[{}\begin{matrix}2x-1=x+2\left(x\ge\dfrac{1}{2}\right)\\1-2x=x+2\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\dfrac{1}{3}\left(tm\right)\end{matrix}\right.\)

Bài 3: 

a: Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}4a+b=3\\2a+b=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a=4\\2a+b=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=-1-2a=-5\end{matrix}\right.\)