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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
Tập hợp A:
\(A=\left\{1;5;9;13;17;21;25\right\}\)
Tập hợp B:
\(B=\left\{0;1;3;5;10;13\right\}\)
Mà: \(A\cap B\)
\(\Rightarrow A\cap B=\left\{1;5;13\right\}\)
⇒ Chọn B
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>25-4x=1
=>4x=24
hay x=6
b: =>2x-4=0
hay x=2
c: =>x-35=115
hay x=150
d: =>x-2=12
hay x=14
e: =>x-36=216
hay x=252
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(=248+2064-12-236\)
\(=12-12+2064=2064\)
b: \(=-298-302-300=-600-300=-900\)
c: \(=5-7+9-11+13-15=-2-2-2=-6\)
d: \(=456+58-456-38=20\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}>4\cdot\dfrac{1}{16}=\dfrac{1}{4}\)
\(\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+\dfrac{1}{20}>4\cdot\dfrac{1}{20}=\dfrac{1}{5}\)
=>\(\dfrac{1}{13}+\dfrac{1}{14}+...+\dfrac{1}{20}>\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{9}{20}\)
=>A>\(\dfrac{1}{12}+\dfrac{9}{20}\)
\(\dfrac{1}{12}>\dfrac{1}{20}\)
=>\(A>\dfrac{1}{20}+\dfrac{9}{20}=\dfrac{1}{2}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{54-34}{189-119}=\dfrac{20}{70}=\dfrac{2}{7}\)
b: \(=\dfrac{6+6\cdot4+6\cdot49}{15+15\cdot4+15\cdot49}=\dfrac{6}{15}=\dfrac{2}{5}\)
c: \(=\dfrac{13\left(3-18\right)}{40\left(15-2\right)}=\dfrac{-15}{40}=-\dfrac{3}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Để $\overrightarrow{a}$ và $\overrightarrow{b}$ cùng phương thì:
\(\frac{2}{5}=\frac{-3}{m}\Rightarrow m=\frac{-15}{2}\)
Đáp án D.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(E=\frac{cosx}{sinx}+\frac{sinx}{1+cosx}=\frac{cosx+cos^2x+sin^2x}{sinx\left(1+cosx\right)}=\frac{cosx+1}{sinx\left(1+cosx\right)}=\frac{1}{sinx}\)
17.
\(\frac{\pi}{2}< a< \pi\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{12}{13}\)
\(0< b< \frac{\pi}{2}\Rightarrow sinb>0\Rightarrow sinb=\sqrt{1-cos^2b}=\frac{4}{5}\)
\(sin\left(a+b\right)=sina.cosb+cosa.sinb=\frac{5}{13}.\frac{3}{5}-\frac{12}{13}.\frac{4}{5}=-\frac{33}{65}\)
18.
\(K=sin\frac{2\pi}{7}+sin\frac{6\pi}{7}+sin\frac{4\pi}{7}\)
\(\Leftrightarrow K.sin\frac{\pi}{7}=sin\frac{\pi}{7}.sin\frac{2\pi}{7}+sin\frac{\pi}{7}.sin\frac{4\pi}{7}+sin\frac{\pi}{7}.sin\frac{6\pi}{7}\)
\(=\frac{1}{2}\left(cos\frac{\pi}{7}-cos\frac{3\pi}{7}+cos\frac{\pi}{7}-cos\frac{5\pi}{7}+cos\frac{5\pi}{7}-cos\frac{7\pi}{7}\right)\)
\(=\frac{1}{2}\left(cos\frac{\pi}{7}-cos\pi\right)=\frac{1}{2}\left(cos\frac{\pi}{7}+1\right)=\frac{1}{2}\left(2cos^2\frac{\pi}{14}-1+1\right)=cos^2\frac{\pi}{14}\)
\(\Leftrightarrow K.2.sin\frac{\pi}{14}.cos\frac{\pi}{14}=cos^2\frac{\pi}{14}\)
\(\Leftrightarrow2K=\frac{cos\frac{\pi}{14}}{sin\frac{\pi}{14}}=cot\frac{\pi}{14}=a\Rightarrow K=\frac{a}{2}\)