dddddddddddddddddd

có thể bỏ ngoạt ko a

Bài 2: 

a, 1/3 + 1/2 : x = -4

=> 1/2 : x = -4 - 1/3 

=> 1/2 : x = -13/3

=> x = 1/2 ; -13/3

=> x = -3/26

Vậy x = -3 / 26

Bài 2: 

b, x² - 4x = 0

=> x.(x - 4) =0

=> x=0 hoặc x - 4 = 0

x - 4= 0 => x=4

Vậy x=0 và x=4

b) \(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2018}\right)\)

\(=\frac{2-1}{2}.\frac{3-1}{3}.\frac{4-1}{4}....\frac{2018-1}{2018}\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2017}{2018}=\frac{1.2.3...2017}{2.3.4...2018}=\frac{1}{2018}\)

c) Giữa các biểu thức là dấu nhân hay dấu cộng vậy bạn?

d)

\(D=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(D=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

e) \(E=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)

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

\(2E=\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+....+\frac{99-97}{97.99}\)

\(2E=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

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

\(\Rightarrow E=\frac{16}{99}\)

#)Well ! Bài này cg dạng tầm cỡ vừa :v 

Làm cho ai v ? mau vô nhận bài đê !

ờ tụi làm cho bạn trên lớp í mà

Mà tui ngu quá nhóm lun từ số 1 nhanh hơn ko 

\(M=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right).....\left(\frac{1}{100}-1\right)\left(\frac{1}{121}-1\right)=\frac{-3}{4}.\frac{-8}{9}.....\frac{-99}{100}.\frac{-120}{121}\)

\(M=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.....\frac{-9.11}{10.10}.\frac{-10.12}{11.11}=\frac{-1}{2}.\frac{-12}{11}=\frac{12}{22}=\frac{6}{11}\)

\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{99}\)

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

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

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

\(S=\frac{5}{11}\)

\(Q=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2013.2015}\)

\(Q=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)

\(Q=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(Q=\frac{1}{2}\left(1-\frac{1}{2015}\right)\)

\(Q=\frac{1007}{2015}\)

~ Đấng Ed :) ~