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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: \(=\dfrac{13\left(3-18\right)}{40\left(15-2\right)}=\dfrac{-15}{40}=-\dfrac{3}{8}\)

17 tháng 8 2023

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

4 tháng 8 2021

Bạn xem hình nha

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2 hình nha hok tốt

4 tháng 8 2021

\(a)13-\left(40-X\right)=35\)

\(\Leftrightarrow40-X=13-35\)

\(\Leftrightarrow40-X=-22\)

\(\Leftrightarrow X=40-\left(-22\right)\)

\(\Leftrightarrow X=62\)

Vậy \(X=62\)

\(b)14-3x\left(5-X\right)=8\)

\(\Leftrightarrow3x\left(5-X\right)=14-8\)

\(\Leftrightarrow3x\left(5-X\right)=6\)

\(\Leftrightarrow5-X=6:3\)

\(\Leftrightarrow5-X=3\)

\(\Leftrightarrow X=5-3\)

\(\Leftrightarrow X=2\)

Vậy \(X=2\)

\(c)\left(3X-2\right)x\left(5+X\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3X-2=0\\5+X=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3X=0+2\\X=0-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3X=2\\X=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}X=2:3\\X=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}X=\frac{2}{3}\\X=-5\end{cases}}\)

Vậy \(X\in\left\{-5;\frac{2}{3}\right\}\)

\(d)\left(3X-6\right)x3=3^4\)

\(\Leftrightarrow\left(3X-6\right)x3=81\)

\(\Leftrightarrow3X-6=81:3\)

\(\Leftrightarrow3X-6=27\)

\(\Leftrightarrow3X=27+6\)

\(\Leftrightarrow3X=33\)

\(\Leftrightarrow X=33:3\)

\(\Leftrightarrow X=11\)

Vậy\(X=11\)

1: \(=\dfrac{1}{29\cdot30}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{28\cdot29}\right)\)

\(=\dfrac{1}{29\cdot30}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{28}-\dfrac{1}{29}\right)\)

\(=\dfrac{1}{29\cdot30}-\dfrac{28}{29}=\dfrac{1-28\cdot30}{870}=\dfrac{-859}{870}\)

B={x\(\in\)N|x=3k; 1<=k<=4}

C={x\(\in\)N|x=4*a2; 1<=a<=5}

D={x\(\in\)N|x=9*a2;1<=a<=4}

E={x\(\in\)N|x=4k; 0<=x<=4}

G={x\(\in\)N|x=(-3)^k; 1<=k<=4}

 

NV
5 tháng 6 2020

\(E=\frac{cosx}{sinx}+\frac{sinx}{1+cosx}=\frac{cosx+cos^2x+sin^2x}{sinx\left(1+cosx\right)}=\frac{cosx+1}{sinx\left(1+cosx\right)}=\frac{1}{sinx}\)

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.

\(K=sin\frac{2\pi}{7}+sin\frac{6\pi}{7}+sin\frac{4\pi}{7}\)

\(\Leftrightarrow K.sin\frac{\pi}{7}=sin\frac{\pi}{7}.sin\frac{2\pi}{7}+sin\frac{\pi}{7}.sin\frac{4\pi}{7}+sin\frac{\pi}{7}.sin\frac{6\pi}{7}\)

\(=\frac{1}{2}\left(cos\frac{\pi}{7}-cos\frac{3\pi}{7}+cos\frac{\pi}{7}-cos\frac{5\pi}{7}+cos\frac{5\pi}{7}-cos\frac{7\pi}{7}\right)\)

\(=\frac{1}{2}\left(cos\frac{\pi}{7}-cos\pi\right)=\frac{1}{2}\left(cos\frac{\pi}{7}+1\right)=\frac{1}{2}\left(2cos^2\frac{\pi}{14}-1+1\right)=cos^2\frac{\pi}{14}\)

\(\Leftrightarrow K.2.sin\frac{\pi}{14}.cos\frac{\pi}{14}=cos^2\frac{\pi}{14}\)

\(\Leftrightarrow2K=\frac{cos\frac{\pi}{14}}{sin\frac{\pi}{14}}=cot\frac{\pi}{14}=a\Rightarrow K=\frac{a}{2}\)

b: \(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{13}{4}=\dfrac{-5}{7}+\dfrac{13}{4}=\dfrac{-20+91}{28}=\dfrac{71}{28}\)

c: \(=\dfrac{146}{13}-3-\dfrac{68}{13}=6-3=3\)

d: \(=\dfrac{2}{7}\left(\dfrac{21}{4}-\dfrac{13}{4}\right)=\dfrac{4}{7}\)

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: \(=456+58-456-38=20\)