E = \(\frac{36}{1\cdot7}+\frac{36}{7\cdot13}+...+\frac{36}{94\cdot100}=\frac{36}{6}\left[\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+...+\frac{1}{94\cdot100}\right]\)

\(=6\left[1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{94}-\frac{1}{100}\right]=6\left[1-\frac{1}{100}\right]\)

\(=6\cdot\frac{99}{100}=\frac{297}{50}\)

F = \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left[3a+2\right]\left[3a+5\right]}\)

\(=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{\left[3a+2\right]\left[3a+5\right]}\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3a+2}-\frac{1}{3a+5}\right]\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3a+5}\right]=\frac{1}{6}-\frac{1}{9a+15}\)

G = \(\frac{1}{2\cdot3}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{4}{8\cdot12}+\frac{5}{12\cdot17}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{12}-\frac{1}{17}\)

\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)

E=36/1-36/7+36/7-36/13+...+36/94-36/100

  =36-36/100=891/25

A = 3 ​(1/3 - 1/5 + 1/5 - 1/8 + 1/8 - 1/12 + 1/12 - 1/17) = 3(1/3 - 1/17) = 14/17

​

​

​

A = \(\frac{6}{3}.5+\frac{9}{5}.8+\frac{12}{8}.12+\frac{15}{12}.17\)

    \(=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}\right)\)

    \(=3\left(\frac{1}{3}-\frac{1}{17}\right)\)

    \(=3\times\frac{14}{51}\)

    \(=\frac{14}{17}\)

CHÚC BẠN HỌC TỐT !!!

\(A=\frac{6}{3.5}+\frac{9}{5.8}+\frac{12}{8.12}+\frac{15}{12.17}\)

\(A=3.\left(\frac{2}{3.5}+\frac{3}{5.8}+\frac{4}{8.12}+\frac{5}{12.17}\right)\)

\(A=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}\right)\)

\(A=3.\left(\frac{1}{3}-\frac{1}{17}\right)< 3.\frac{1}{3}=1\)

=> A < 1

Ta có :

\(A=\frac{6}{3.5}+\frac{9}{5.8}+\frac{12}{8.12}+\frac{15}{12.17}\)

\(A=3.\left(\frac{2}{3.5}\right)+3.\left(\frac{3}{5.8}\right)+3.\left(\frac{4}{8.12}\right)+3.\left(\frac{5}{12.17}\right)\)

\(A=3.\left(\frac{2}{3.5}+\frac{3}{5.8}+\frac{4}{8.12}+\frac{5}{12.17}\right)\)

\(A=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}\right)\)

\(A=3.\left(\frac{1}{3}-\frac{1}{17}\right)\)

\(A=3.\frac{14}{51}\)

\(A=\frac{14}{17}< 1\)

Vậy A < 1

_Chúc bạn học tốt_

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

~ Hok tốt ~

\(\)

Viết thành 2 . (1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/97.99

a: ta có: \(5\cdot8^{12}\cdot9^5\cdot\left(-3\right)^2\cdot\left(-16\right)^5\)

\(=5\cdot\left(2^3\right)^{12}\cdot\left(3^2\right)^5\cdot3^2\cdot\left(-1\right)\cdot2^{20}\)

\(=-5\cdot3^{12}\cdot2^{56}\)

Ta có: \(7\cdot49^3\cdot81^5\cdot\left(-7\right)^{10}\cdot\left(-8\right)^5\)

\(=-7\cdot\left(7^2\right)^3\cdot\left(3^4\right)^5\cdot7^{10}\cdot\left(2^3\right)^5\)

\(=-7\cdot7^6\cdot3^{20}\cdot7^{10}\cdot2^{15}=-7^{17}\cdot2^{15}\cdot3^{20}\)

Ta có: \(B=\frac{5\cdot8^{12}\cdot9^5\cdot\left(-3\right)^2\cdot\left(-16\right)^5}{7\cdot49^3\cdot81^5\cdot\left(-7\right)^{10}\cdot\left(-8\right)^5}\)

\(=\frac{-5\cdot3^{12}\cdot2^{56}}{-7^{17}\cdot2^{15}\cdot3^{20}}=\frac{5\cdot2^{41}}{7^{17}\cdot3^8}\)

b: \(13\cdot4^7\cdot9^{15}-9^7\cdot\left(-16\right)^4\)

\(=13\cdot2^{14}\cdot3^{20}-3^{14}\cdot2^{16}\)

\(=3^{14}\cdot2^{14}\left(13\cdot3^6-2^2\right)\)

\(15\cdot3^{12}\cdot2^8+3\cdot27^5\cdot\left(-8\right)^{10}\)

\(=3\cdot5\cdot3^{12}\cdot2^8+3\cdot\left(3^3\right)^5\cdot\left(2^3\right)^{10}=3^{13}\cdot5\cdot2^8+3^{16}\cdot2^{30}\)

\(=3^{13}\cdot2^8\left(5+3^3\cdot2^{22}\right)\)
Ta có: \(C=\frac{13\cdot4^7\cdot9^{15}-9^7\cdot\left(-16\right)^4}{15\cdot3^{12}\cdot2^8+3\cdot27^5\cdot\left(-8\right)^{10}}\)

\(=\frac{3^{14}\cdot2^{14}\left(13\cdot3^6-2^2\right)}{3^{13}\cdot2^8\left(5+3^3\cdot2^{22}\right)}=\frac{3\cdot2^6\cdot\left(13\cdot3^6-2^2\right)}{5+3^3\cdot2^{22}}\)

Câu 2:

\(D=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)

Câu 3: 

\(E=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{205}-\dfrac{1}{207}\right)\)

\(=2\cdot\left(1-\dfrac{1}{207}\right)=2\cdot\dfrac{206}{207}=\dfrac{412}{207}\)

Câu 5: 

\(G=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{16}{17}=\dfrac{4}{17}\)

\(\frac{1.2+1.4+3.6+4.8}{2.3+4.6+6.9+8.12}\)

=\(\frac{1.2}{2.3}\)+\(\frac{1.4}{4.6}\)+\(\frac{3.6}{6.9}\)+\(\frac{4.8}{8.12}\)

= \(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{3}\)+\(\frac{1}{3}\)

= \(\frac{2}{6}+\frac{1}{6}+\frac{2}{6}+\frac{2}{6}\)

=\(\frac{7}{6}\)

Mình nghĩ đề bài phải là:

          \(\frac{1.2+2.4+3.6+4.8}{2.3+4.6+6.9+8.12}\)   *2.3 + 4.6 + 6.9 + 8.12 = 3.(1.2 + 2.4 + 3.6 + 4.8)*

\(=\)\(\frac{1\left(1.2+2.4+3.6+4.8\right)}{3\left(1.2+2.4+3.6+4.8\right)}\)

\(=\)\(\frac{1}{3}\)

đề kiểu gì vậy bn

(2,x+10) á

xàm lone :))