19.C

20.D

21.A

22.B

23.C

24.B

26.C

28.D

31D

32A

33C

29:

\(sinB=\sqrt{1-\dfrac{1}{64}}=\sqrt{\dfrac{63}{64}}=\dfrac{3\sqrt{7}}{8}\)

\(sinC=\sqrt{1-\dfrac{9}{16}}=\sqrt{\dfrac{7}{16}}=\dfrac{\sqrt{7}}{4}\)

\(sin\left(B+C\right)=sinB\cdot cosC+sinC\cdot cosB\)

\(=\dfrac{3\sqrt{7}}{8}\cdot\dfrac{3}{4}+\dfrac{1}{8}\cdot\dfrac{\sqrt{7}}{4}=\dfrac{10\sqrt{7}}{32}=\dfrac{5\sqrt{7}}{16}\)

=>\(sinA=\dfrac{5\sqrt{7}}{16}\)

\(\dfrac{BC}{sinA}=\dfrac{AC}{sinB}\)

=>\(BC:\dfrac{5\sqrt{7}}{16}=6:\dfrac{3\sqrt{7}}{8}=\dfrac{48}{3\sqrt{7}}=\dfrac{16}{\sqrt{7}}\)

=>\(BC=\dfrac{16}{\sqrt{7}}\cdot\dfrac{5\sqrt{7}}{16}=5\)

Ta có:

\(\left\{{}\begin{matrix}GM=\dfrac{1}{3}AM=5\\GN=\dfrac{1}{3}BN=4\end{matrix}\right.\)

\(S_{CMN}=S_{AMN}=15\sqrt{3}\)

\(S_{GMN}=\dfrac{1}{3}S_{AMN}=5\sqrt{3}\)

=> \(5\sqrt{3}=\dfrac{1}{2}.GM.GN.sinG\)

\(\Rightarrow sinG=\dfrac{\sqrt{3}}{2}\Rightarrow\widehat{G}=60^o\)

\(MN=\sqrt{GM^2+GN^2-2GM.GN.cosG}=\sqrt{21}\)

Làm ơn giúp en em sẽ đánh đúng cho người đầu tiên

a) \(\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{IB}-\overrightarrow{IB}=\overrightarrow{0}\)

b) \(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\)

\(=2\overrightarrow{GE}+\overrightarrow{GC}=2\overrightarrow{GE}-2\overrightarrow{GE}=\overrightarrow{0}\)

Ta có:

\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\) (tính chất tổng 3 góc trong 1 tam giác)

\(\Rightarrow\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{2}=90^o\)

\(\Rightarrow\dfrac{\widehat{B}+\widehat{C}}{2}=90^o-\dfrac{\widehat{A}}{2}\)

\(\Rightarrow\)\(tan\left(\dfrac{\widehat{B}+\widehat{C}}{2}\right)=tan\left(90^o-\widehat{\dfrac{A}{2}}\right)\)

\(\Rightarrow tan\left(\dfrac{\widehat{B}+\widehat{C}}{2}\right)=cot\dfrac{A}{2}\)

Lời giải:
Để $A\cap B=\varnothing$ thì:

\(\left[\begin{matrix} 2m+1<-1\\ 2m-1\geq 5\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} m<-1\\ m\geq 3\end{matrix}\right.\)

Đáp án B.

$A\cup B=A\Leftrightarrow$ \(A\supset B\)