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26 tháng 4 2020

Tách ra là xong nhé!!

S=1/2-1/100=49/100

P=1-1/94=93/94

k mình đúng với!!!!

3 tháng 5 2019

Ta có: \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}\)

\(\Leftrightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{94}-\frac{1}{97}\)

\(\Leftrightarrow1-\frac{1}{97}=\frac{96}{97}\)

Do \(\frac{96}{97}< 1\Rightarrow\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.97}< 1\)

Vậy:.............................<1

21 tháng 3 2018

c)1*(1/2-1/3+1/3-1/4+.....+1/91-1/94)

1/2-1/94 ban tu tinh nhe

d)1*(1/1-1/4+1/4-1/7+......+1/91-1/94)

1-1/94 ban tu tinh nhe 

tk nha

21 tháng 3 2018

a) \(\frac{1}{n}-\frac{1}{n+1}\left(n\inℕ^∗\right)\)

\(\Leftrightarrow\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}\Leftrightarrow\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)

b) \(\frac{1}{n}-\frac{1}{n+3}\left(n\inℕ^∗\right)\)

\(\Leftrightarrow\frac{n+3}{n\left(n+3\right)}-\frac{n}{n\left(n+3\right)}=\frac{n+3-n}{n\left(n+3\right)}=\frac{3}{n\left(n+3\right)}\)

c,d dễ bn tách ra rồi trừ đi

27 tháng 7 2017

\(A=\frac{3}{1.4}+\frac{3}{4.7}+..........+\frac{3}{91.94}\)

\(\Leftrightarrow A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{91}-\frac{1}{94}\)

\(\Leftrightarrow A=1-\frac{1}{94}=\frac{93}{94}\)

\(B=\frac{1}{1.3}+\frac{1}{3.5}+......+\frac{1}{97.99}\)

\(\Leftrightarrow2B=\frac{2}{1.3}+\frac{2}{3.5}+.......+\frac{3}{97.99}\)

\(\Leftrightarrow2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{97}-\frac{1}{99}\)

\(\Leftrightarrow2B=1-\frac{1}{99}=\frac{98}{99}\)

\(\Leftrightarrow B=\frac{98}{99}:2=\frac{49}{99}\)

27 tháng 7 2017

Ta có : \(A=\frac{3}{1.4}+\frac{3}{4.7}+.....+\frac{3}{91.94}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{91}-\frac{1}{94}\)

\(=1-\frac{1}{94}\)

\(=\frac{93}{94}\)

26 tháng 4 2018

\(\frac{3}{1.4}+\frac{3}{4.7}+.....+\frac{3}{94.97}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.........+\frac{1}{94}-\frac{1}{97}\)

\(=1-\frac{1}{97}\)

\(=\frac{96}{97}\)

mà \(\frac{96}{97}< 1\)

\(\Rightarrow\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.07}< 1\)

vậy..................

\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{91\cdot94}+\frac{3}{94\cdot97}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\)

\(=1-\frac{1}{97}\)

\(=\frac{96}{97}\)

\(\Rightarrow\frac{96}{97}< 1\)

\(\Rightarrow\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{94\cdot97}< 1\)

Vậy \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{94\cdot97}< 1\)

a: =1/2-1/3+1/3-1/4+...+1/99-1/100

=1/2-1/100=49/100

b; =5/3(1-1/4+1/4-1/7+...+1/100-1/103)

=5/3*102/103

=510/309=170/103

c: =1/2(1/3-1/5+1/5-1/7+...+1/49-1/51)

=1/2*16/51=8/51

6 tháng 3 2023

\(B=1-\dfrac{3}{1\cdot4}-\dfrac{3}{4\cdot7}-...-\dfrac{3}{2020\cdot2023}\\ =1-\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2020\cdot2023}\right)\\ =1-\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2020}-\dfrac{1}{2023}\right)\\ =1-\left(1-\dfrac{1}{2023}\right)\\ =1-\dfrac{2022}{2023}=\dfrac{1}{2023}\)

6 tháng 3 2023

`B=1-3/(1.4)-3/(4.7)-3/(7.10)-....-3/(2020.2023)`

`B=1-(3/(1.4)+3/(4.7)+.....+3/(2020.2023))`

`B=1-(1-1/4+1/4-1/7+.....+1/2020-1/2023)`

`B=1-(1-1/2023)`

`B=1-1+1/2023=1/2023`

8 tháng 5 2023

A = \(\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+...+\dfrac{3^2}{196.199}\)

A = \(\dfrac{3.3}{1.4}+\dfrac{3.3}{4.7}+...+\dfrac{3.3}{196.199}\)

A = \(3.\dfrac{3}{1.4}+3.\dfrac{3}{4.7}+...+3.\dfrac{3}{196.199}\)

A = \(3\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{196.199}\right)\)

A = \(3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{196}-\dfrac{1}{199}\right)\)

A = \(3\left(1-\dfrac{1}{199}\right)\) = \(3.\dfrac{198}{199}\) = \(\dfrac{594}{199}\)

9 tháng 4 2017

a)\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{46}-\frac{1}{56}\)

=\(1-\frac{1}{56}=\frac{55}{56}\)

b)\(A.\frac{1}{3}=\frac{1}{3}.\left(\frac{3}{1.2}+\frac{3}{2.3}+....+\frac{3}{99.100}\right)\)

\(\frac{1}{3}A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{3}{99.100}\)

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

=> \(\frac{1}{3}A=1-\frac{1}{100}=\frac{99}{100}\)

=> \(A=\frac{99}{100}.3=\frac{297}{100}\)

c)\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)

=\(1-\frac{1}{103}=\frac{102}{103}\)

d) \(\frac{3}{5}C=\frac{3}{5}.\left(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\right)\)

=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\)

=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\)

=\(1-\frac{1}{103}=\frac{102}{103}\)

=>\(C=\frac{102}{103}.\frac{5}{3}=\frac{170}{103}\)

e) \(\frac{4}{7}D=\frac{4}{7}.\left(\frac{7}{1.5}+\frac{7}{5.9}+...+\frac{7}{101.105}\right)\)

=\(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{101.105}\)

=\(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)

=\(1-\frac{1}{105}=\frac{104}{105}\)

=< D=\(\frac{104}{105}.\frac{7}{4}=\frac{26}{15}\)