\(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{13\cdot15}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{4}{15}=\dfrac{2}{15}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{5}{15}-\dfrac{1}{15}=\dfrac{4}{15}\)

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\)

= \(\dfrac{1}{3}-\dfrac{1}{13}\)

=\(\dfrac{10}{26}=\dfrac{5}{13}\)

1100444-88888=

a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{1}{2}.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\)

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

\(=\frac{1}{2}.\left(1-\frac{1}{11}\right)\)

\(\frac{10}{22}\)

thế phần b và c đâu???

\(\dfrac{12}{1.4.7}+\dfrac{12}{4.7.10}+\dfrac{12}{7.10.13}+...+\dfrac{12}{54.57.60}\)

\(=2\left(\dfrac{1}{1.4}-\dfrac{1}{4.7}+\dfrac{1}{4.7}-\dfrac{1}{7.10}+\dfrac{1}{7.10}-\dfrac{1}{10.13}+...+\dfrac{1}{54.57}-\dfrac{1}{57.60}\right)\)\(=2\left(\dfrac{1}{1.4}-\dfrac{1}{57.60}\right)\)

\(=2\left(\dfrac{1}{4}-\dfrac{1}{57.60}\right)=\dfrac{1}{2}-\dfrac{1}{2.57.60}< \dfrac{1}{2}\left(đpcm\right)\)

`a)1/2 . [-3]/4 . [-5]/8 . [-8]/9=[1. (-3).(-5).(-8)]/[2.4.8.3.3]=[-5]/[2.4.3]=[-5]/24`

`b)(2/[1.3]+2/[3.5]+2/[5.7]).([10.13]/3-[2^2]/3-[5^3]/3)`

`=(1-1/3+1/3-1/5+1/5-1/7).[10.13-2^2-5^3]/3`

`=(1-1/7).[130-4-125]/3`

`=6/7 . 1/3 = 2/7`

____________________________________________________

`8/9+1/9 . 2/9+1/9 . 7/9`

`=8/9+1/9.(2/9+7/9)`

`=8/9+1/9 . 9/9`

`=8/9+1/9=9/9=1`

a) ^{_{\(\dfrac{1}{2}\cdot\dfrac{-3}{4}\cdot\dfrac{-5}{8}\cdot\dfrac{-8}{9}\)}}

^{_{\(=\dfrac{1\cdot\left(-3\right)\cdot\left(-5\right)\cdot\left(-8\right)}{2\cdot4\cdot8\cdot9}\)}}

^{_{\(=-\dfrac{5}{24}\)}}

b) ^{_{\(\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}\right)\cdot\left(\dfrac{10\cdot13}{3}-\dfrac{2^2}{3}-\dfrac{5^3}{3}\right)\)}}

^{_{\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}\right)\cdot\left(\dfrac{130}{3}-\dfrac{4}{3}-\dfrac{125}{3}\right)\)}}

_{^{\(=\left(1-\dfrac{1}{7}\right)\cdot\dfrac{1}{3}\)}}

_{^{\(=\dfrac{6}{7}\cdot\dfrac{1}{3}\)}}

_{^{\(=\dfrac{2}{7}\)}}

_{^{\(\dfrac{8}{9}+\dfrac{1}{9}\cdot\dfrac{2}{9}+\dfrac{1}{9}\cdot\dfrac{7}{9}\)}}

_{^{\(=\dfrac{8}{9}+\dfrac{2}{81}+\dfrac{7}{81}\)}}

_{^{\(=\dfrac{72}{81}+\dfrac{2}{81}+\dfrac{7}{81}\)}}

_^\(=1\)

ta có: \(\frac{2^5.7+2^5}{2^5.2^5-2^5.3}=\frac{2^5.\left(7+1\right)}{2^5.\left(2^5-3\right)}=\frac{8}{2^5-3}=\frac{8}{29}=\frac{104}{377}\)

\(\frac{3^4.5.\left(-3\right)^6}{3^4.13.3^4}=\frac{3^{10}.5}{3^8.13}=\frac{3^2.5}{13}=\frac{45}{13}=\frac{1305}{377}\)

\(\Rightarrow\frac{104}{377}< \frac{1305}{377}\Rightarrow\frac{2^5.7+2^5}{2^5.2^5-2^5.3}< \frac{3^4.5.\left(-3\right)^6}{3^4.13.3^4}\)

Ta cứ tính ra tử số và mỗi số của từng phân số ra nhé Jerry Gaming:

\(\frac{2^5.7+2^5}{2^5.2^5-2^5.3}\)= \(\frac{2^5.\left(7+1\right)}{2^5.\left(2^5-3\right)}=\frac{2^5.8}{2^5.\left(32-3\right)}=\frac{32.8}{2^5.29}=\frac{32.8}{32.29}=\frac{8}{29}\)

\(\frac{3^4.5.\left(-3\right)^6}{3^4.13.3^4}\)= \(\frac{3^4.5.3^6}{3^8.13}=\frac{3^{10}.5}{3^8.13}=\frac{3^2.5}{13}=\frac{9.5}{13}=\frac{45}{13}\)

\(\frac{8}{29}\)và \(\frac{45}{13}\)MSC: 377

Ta có:

\(\frac{8}{29}=\frac{8.13}{29.13}=\frac{104}{377}\)

\(\frac{45}{13}=\frac{45.29}{13.29}=\frac{1305}{377}\)

Vậy quy đồng \(\frac{2^5.7+2^5}{2^5.2^5-2^5.3}\)và \(\frac{3^4.5.\left(-3\right)^6}{3^4.13.3^4}\)ta được \(\frac{104}{377}\)và \(\frac{1305}{377}\)

Chúc bạn học tốt!

\(\frac{2^{10}\cdot13+2^{10}\cdot37}{2^8\cdot90}\)

\(=\frac{2^{10}\cdot\left(13+37\right)}{2^8\cdot2\cdot3^2\cdot5}\)

\(=\frac{2^{10}\cdot50}{2^9\cdot3^2\cdot5}\)

\(=\frac{2^{10}\cdot2\cdot5^2}{2^9\cdot3^2\cdot5}=\frac{2^{11}\cdot5^2}{2^9\cdot3^2\cdot5}\)

\(=\frac{2^2\cdot5}{3^2}=\frac{20}{9}\)

\(\frac{2^{10}.13+2^{10}.27}{2^8.90}\)

=\(\frac{2^{10}.\left(13+27\right)}{2^8.90}\)

=\(\frac{2^{10}.40}{2^8.90}\)

=\(\frac{2^2.4}{1.9}\)

=\(\frac{4.4}{9}\)

=\(\frac{16}{9}\)

\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)

\(=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\right)\)

\(=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)

\(=\dfrac{1}{3}-\dfrac{1}{13}\)

\(=\dfrac{10}{39}\)