Ta có : (ghi đầu bài)

           =10101*(5/111111+5/222222-4/111111)

           =10101*5/111111+10101*5/222222-10101*4/111111 (nhân phân phối)

           =5/11+5/22-4/11

           =10/22+5/22-8/22

           =7/22

Chúc May MẮn

a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\)

\(=\frac{1}{2}-\frac{1}{3n+2}=\frac{3n+2}{2\cdot\left(3n+2\right)}-\frac{2}{2\cdot\left(3n+2\right)}\)

\(=\frac{3n+2-2}{6n+4}=\frac{3n}{6n+4}=VP\)

chết phần a quên nhân vs 1/3

Aps dụng tính chất phân phối của phép nhân đối với phép cộng,ta được:

\(\frac{5}{11}+\frac{5}{22}-\frac{4}{11}=\frac{7}{22}\)

#Châu's ngốc

a) = \(\dfrac{12}{17}\)

b) = \(\dfrac{31}{37}\)

2/ 

a) \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)

\(=\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{17}-\frac{1}{21}\right)\)

\(=1-\frac{1}{21}=\frac{20}{21}\)

b) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot..\cdot\frac{2016}{2017}\)

\(=\frac{1}{2017}\)

c) \(A=2000-5-5-5-..-5\)(có 200 số 5) 

\(A=2000-\left(5\cdot200\right)\)

\(A=2000-1000\)

\(A=1000\)

\(a)\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)

\(b)\frac{-11.13^7}{11^5.13^8}=\frac{-1}{11^4.13}\) (Bạn xem thử xem có sai đề không nhé)

\(c)\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3+1\right)}{2^9.3^{10}}=\frac{2.4}{3}=\frac{8}{3}\)

\(d)\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(5.4+9\right)}=\frac{8}{20+9}=\frac{8}{29}\)

\(a)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)

\(b)\frac{\left(-11\right)\cdot13^7}{11^5\cdot13^8}=\frac{-1}{11^4\cdot13}=\frac{-1}{14641\cdot13}=\frac{-1}{190333}\)

\(c)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\left(3^{10}-3^9\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\cdot2}{2^9\cdot3^{10}}=\frac{2\cdot2}{3}=\frac{4}{3}\)

SAI HẾT RỒI.........CẦN THÌ TỚ GIẢI LẠI CHO !!

thế này :

= \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{11.13}\right)\)

= \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{11}-\frac{1}{13}\right)\)

= \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)

= \(\frac{1}{2}.\frac{10}{39}\)

=  \(\frac{5}{39}\)

Vậy kq = \(\frac{5}{39}\)

\(\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)\times y=\frac{2}{3}\)

\(\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\times y=\frac{2}{3}\)

\(\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{11}\right)\times y=\frac{2}{3}\)

\(\frac{1}{2}\times\frac{10}{11}\times y=\frac{2}{3}\)

\(\frac{5}{11}\times y=\frac{2}{3}\) => \(y=\frac{2}{3}:\frac{5}{11}=\frac{2}{3}\times\frac{11}{5}=\frac{22}{15}\)