\(B=\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot\left(1-\dfrac{1}{10}\right)\cdot\left(1-\dfrac{1}{15}\right)\cdot...\cdot\left(1-\dfrac{1}{780}\right)\)

\(B=\left(1-\dfrac{1}{3+6+10+15+...+780}\right)\)

\(B=\left(1-\dfrac{1}{\left(780-3\right)\div3+1}\right)\)

\(B=\left(1-\dfrac{1}{260}\right)\)

\(B=\dfrac{259}{260}\)

Ta có: \(\dfrac{-5}{13}\cdot\dfrac{3}{7}-\dfrac{2}{17}\cdot\dfrac{8}{13}+\dfrac{5}{13}\cdot\dfrac{1}{7}\)

\(=\dfrac{5}{13}\left(-\dfrac{3}{7}+\dfrac{1}{7}\right)-\dfrac{2}{17}\cdot\dfrac{8}{13}\)

\(=\dfrac{5}{13}\cdot\dfrac{-2}{7}-\dfrac{2}{17}\cdot\dfrac{8}{13}\)

\(=\dfrac{-10}{91}-\dfrac{16}{221}\)

\(=\dfrac{-282}{1547}\)

cảm ơn bạn nhé

good chị nhất luôn

\(\Rightarrow y\times\left(2+\dfrac{1}{5}\right)=\dfrac{8}{5}\\ \Rightarrow y\times\dfrac{11}{5}=\dfrac{8}{5}\\ \Rightarrow y=\dfrac{8}{5}:\dfrac{11}{5}=\dfrac{8}{5}\times\dfrac{5}{11}=\dfrac{8}{11}\)

⇔\(x\) x 1=\(\dfrac{8}{3}\)

⇔\(x\) =\(\dfrac{8}{3}\)

Vậy \(x\)=\(\dfrac{8}{3}\)

mik ko hiểu ?

\(\Rightarrow x+\dfrac{1}{6}=\dfrac{3}{4}\\ \Rightarrow x=\dfrac{7}{12}\)

X+

Bài 1: 

=>y-69,3=20,5

hay y=89,8

Bài 2: 

a: \(=365.4:43.5-7.7\cdot0.2\)

=8,4-1,54=6,86

b: \(=\dfrac{2}{5}:\dfrac{8-5}{10}+\dfrac{3}{5}=\dfrac{2}{5}\cdot\dfrac{10}{3}+\dfrac{3}{5}=\dfrac{20+9}{15}=\dfrac{29}{15}\)

Sửa đề: ΔABC vuông tại A

a)Sửa đề: C/m ΔHBA\(\sim\)ΔABC

Xét ΔHBA vuông tại H và ΔABC vuông tại A có 

\(\widehat{B}\) chung

Do đó: ΔHBA\(\sim\)ΔABC(g-g)

b) Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:

\(BC^2=AB^2+AC^2\)

\(\Leftrightarrow BC^2=12^2+10^2=244\)

hay \(BC=2\sqrt{61}cm\)

Vậy: \(BC=2\sqrt{61}cm\)