bạn xem lại đề bài có đúng không nhé. Hình như chỗ -1/4 phải là 1/4 mới đúng

=(-3042014/152014+152015/3042015)*(-7/12+4/122+3/12)

=(-3042014/152014+152015/3042015)*0

=0

(-3042014\152014+152015\3042015).(-7\12+4\12_-3\12)

=>(-3042014\152014+152015\3042015).0

=>0

Ta có :

a) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)

\(=\)\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)

\(=\)\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\)\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)\)

\(=\)\(\frac{1}{2}.\frac{4}{15}\)

\(=\)\(\frac{2}{15}\)

Ta có : 

\(c)\)\(\frac{1}{1000}+\frac{13}{1000}+\frac{25}{1000}+\frac{37}{1000}+...+\frac{229}{1000}\)

\(=\)\(\frac{1+13+25+37+...+229}{1000}\)

Xét tổng \(1+13+25+37+...+229\):

Số số hạng : \(\left(229-1\right):12+1=20\) ( số hạng )

Tổng : \(\frac{\left(229+1\right).20}{2}=2300\)

Do đó :

\(\frac{1+13+25+37+...+229}{1000}=\frac{2300}{1000}=\frac{23}{10}\)

Tính nhanh hả bạn

a) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)

=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)

=\(1-\dfrac{1}{6}\)=\(\dfrac{5}{6}\)

b) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)

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

=\(\dfrac{1.2}{3.5.2}+\dfrac{1.2}{5.7.2}+\dfrac{1.2}{7.9.2}+\dfrac{1.2}{9.11.2}+\dfrac{1.2}{11.13.2}\)

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

=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)

=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)=\(\dfrac{1}{2}.\dfrac{10}{39}\)=\(\dfrac{5}{39}\).

c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)

=\(1-\dfrac{1}{8}=\dfrac{7}{8}\).

d) \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)

=\(\dfrac{2^4}{2^5}+\dfrac{2^3}{2^5}+\dfrac{2^2}{2^5}+\dfrac{2}{2^5}+\dfrac{1}{2^5}\)

=\(\dfrac{2^4+2^3+2^2+2+1}{2^5}\)=\(\dfrac{2^5-1}{2^5}=\dfrac{31}{32}\).

e) \(\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}=\dfrac{7^{99}+7^{98}+7^{97}+...+7+1}{7^{100}}=\dfrac{\dfrac{7^{100}-1}{6}}{7^{100}}=\dfrac{7^{100}-1}{6.7^{100}}\)

\(\left(\frac{-3042014}{152014}-\frac{152014}{3042014}\right)\left(\frac{7}{12}+\frac{1}{3}-\frac{1}{4}\right)\)

\(=\left(-200,1298189\right)\left(\frac{7}{12}+\frac{4}{12}-\frac{3}{12}\right)\)

\(=\left(-200,1298189\right).\frac{2}{3}\)

\(=-133,4198793\)

chúc bạn học tốt

\(\left(\frac{-3042014}{152014}-\frac{152014}{3042014}\right)\left(\frac{-7}{12}+\frac{1}{3}-\frac{-1}{4}\right)\)

\(=\left(\frac{-3042014}{152014}-\frac{152014}{3042014}\right)\left(\frac{-7}{12}+\frac{1}{3}+\frac{1}{4}\right)\)

\(=\left(\frac{-3042014}{152014}-\frac{152014}{3042014}\right)\left(\frac{-7}{12}+\frac{7}{12}\right)\)

\(=\left(\frac{-3042014}{152014}-\frac{152014}{3042014}\right).0=0\)

N=7/2(2/1.3+....+2/13.15)

N=7/2.(1/1-1/3+.....+1/13-1/15)

N=7/2.(1-1/15)

N=7/2.(14/15)

N=7.14/2.15

Bạn giải M luôn dc ko?

Dễ lắm bạn.Suy nghĩ 1 tí là ra

mình giải từng bài nhá

hả đơn giản

4.x=16