Trả lời :

\(\frac{42}{15.7}\)= \(\frac{2.3.7}{3.5.7}\)= \(\frac{2}{5}\)

\(\frac{4.33}{11.12}\)= \(\frac{4.3.11}{11.3.4}\)= \(1\)

_Học tốt

\(\frac{42}{15.7}=\frac{3.2.7}{3.5.7}=\frac{2}{5}\)

\(\frac{36.6}{336}=\frac{3.4.3.3.2}{2.4.2.3.7}=\frac{3.3}{2.7}=\frac{9}{14}\)

\(\frac{4.33}{11.12}=\frac{4.11.3}{11.3.4}=1\)

Trả lời :

\(E=-\left(\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{n\left(n+4\right)}\right)\)

\(\Rightarrow E=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n}-\frac{1}{n+4}\right)\)

\(\Rightarrow E=-\left(1-\frac{1}{n+4}\right)\)

\(\Rightarrow E=1+\frac{1}{n+4}\)

P/s : Sai thì thông cảm nha chị. Dạng này lâu chưa làm nên không nhớ rõ.

\(E=-\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.11}-...-\frac{4}{\left(n-4\right)n}\)

\(\Rightarrow E=-\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.11}+...+\frac{4}{\left(n-4\right)n}\right)\)

\(\Rightarrow E=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)

\(\Rightarrow E=-\left(1-\frac{1}{n}\right)\)

\(\Rightarrow E=-1+\frac{1}{n}\)

a ) \(\frac{3^2.5^2+15.7}{22.6+18.11}\)

= \(\frac{15.15+15.7}{33.4+6.33}\)

= \(\frac{15.\left(15+7\right)}{33.\left(4+6\right)}\)

= \(\frac{15.22}{33.10}\)

= \(\frac{3.5.11.2}{11.3.5.2}\)

= 1

a,\(\frac{3^2.5^2+15.7}{22.6+18.11}=\frac{9.15+15.7}{11.12+18.11}=\frac{\left(9+7\right).15}{11\left(12+18\right)}\)\(=\frac{16.15}{11.30}=\frac{8}{11}\)

b,\(\frac{6.9-2.17}{63.3-119}=\frac{2.27-2.17}{7.27-7.17}=\frac{2.\left(27-17\right)}{7.\left(27-17\right)}\)\(=\frac{2}{7}\)

c,\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.16}=\frac{3.16}{16}=3\)

bn ơi, cho mjk cách lm với 

\(A=\frac{10.11+50.55+70.77}{11.12+55.60+77.84}\)

\(=\frac{10.11+5.10.5.11+7.10.7.11}{11.12+11.5.12.5+11.7.12.7}\)

\(=\frac{10.11\left(1+25+49\right)}{11.12\left(1+25+49\right)}\)

\(=\frac{10.11}{11.12}=\frac{10}{12}=\frac{5}{6}\)

\(B=\frac{1\times3\times5\times7\times........\times49}{26\times27\times28\times...........\times50}\)

\(=\frac{\left(1\times3\times5\times7\times.........\times49\right).\left(2\times4\times6.........48\times50\right)}{\left(26\times27\times28\times.........\times50\right).\left(2\times4\times6\times...........\times48\times50\right)}\)

\(=\frac{1\times2\times3\times4\times..........\times50}{\left(26\times27\times28\times..............\times50\right)2^{25}\left(1\times2\times3\times4\times............\times25\right)}=\frac{1}{2^{25}}\)

\(C=\frac{1.2.6+2.4.12+4.8.24+7.14.42}{1.6.9+2.12.18+4.24.36+7.42.63}\)

\(=\frac{1.2.6\left(1+8+64+343\right)}{1.6.9\left(1+8+64+343\right)}\)

\(=\frac{1.2.6}{1.6.9}=\frac{2}{9}\)

\(A=\frac{5}{6}\)

\(B=\frac{1}{33554432}\)

\(C=\frac{28}{117}\)

\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+\frac{2}{15.17}\)

\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{17}\)

\(A=1-\frac{1}{17}\)

\(A=\frac{16}{17}\)

\(B=\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{9.11}+\frac{4}{11.13}\)

\(B=\frac{4}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(B=\frac{4}{2}\left(1-\frac{1}{13}\right)\)

\(B=\frac{4}{2}\cdot\frac{12}{13}\)

\(B=\frac{24}{13}\)

=> A= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}\)

=> A= \(\frac{1}{1}-\frac{1}{17}\)

=> A= \(\frac{16}{17}\)

\(\Rightarrow B=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(\Rightarrow B=2.\left(\frac{1}{1}-\frac{1}{13}\right)\)

\(\Rightarrow B=2.\frac{12}{13}\)

\(\Rightarrow B=\frac{24}{13}\)

Tổng quát: \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\) (với mọi số tự nhiên n khác 0)

Ta có: \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}<\frac{1}{2}\) (vì \(\frac{1}{100}>0\) )

=>đpcm

\(\frac{10.11+50.55+70.77}{11.12+55.60+77.84}=\frac{10.11.\left(1+5.5+7.7\right)}{11.12.\left(1+5.5+7.7\right)}=\frac{10}{12}=\frac{5}{6}\)