a.\(\dfrac{27}{8}\)

b.\(\dfrac{37}{40}\)
c.\(\dfrac{5}{2}\)

d.\(\dfrac{7}{3}\)

e.5

g.\(\dfrac{53}{16}\)

Bài 1 :

a) \(\dfrac{3}{2}+\dfrac{5}{4}+\dfrac{5}{8}=\dfrac{12}{8}+\dfrac{10}{8}+\dfrac{5}{8}=\dfrac{12+10+5}{8}=\dfrac{27}{8}\)

b) \(\dfrac{4}{5}-\dfrac{3}{8}+\dfrac{2}{4}=\dfrac{32}{40}-\dfrac{15}{40}+\dfrac{20}{40}=\dfrac{32-15+20}{40}=\dfrac{37}{40}\)

c) \(3+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{3}{1}+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{24}{8}+\dfrac{6}{8}-\dfrac{10}{8}=\dfrac{20}{8}=\dfrac{5}{2}\)

d) \(\dfrac{5}{6}-\dfrac{1}{2}+2=\dfrac{5}{6}-\dfrac{1}{2}+\dfrac{2}{1}=\dfrac{5}{6}-\dfrac{3}{6}+\dfrac{12}{6}=\dfrac{14}{6}=\dfrac{7}{3}\)

e) \(\dfrac{3}{5}+\dfrac{6}{11}+\dfrac{7}{13}+\dfrac{2}{5}+\dfrac{16}{11}+\dfrac{19}{13}=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{6}{11}+\dfrac{16}{11}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)=1+2+2=5\)

g) \(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}=\dfrac{3}{4}+\dfrac{6}{7}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{13}{32}=\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{6}{7}+\dfrac{1}{7}\right)+\left(\dfrac{29}{32}+\dfrac{13}{32}\right)=1+1+\dfrac{21}{16}=2+\dfrac{21}{16}=\dfrac{53}{16}\)

a,3/2+5/4+5/8 = 12/8 + 10/8 + 5/8 = 27/8

b,4/5-3/8+2/4 = 32/40 - 15/40 + 20/40 = 37/40

c,3+6/8-5/4= 24/8 + 6/8 - 10/8 = 20/8 = 5/2

d,5/6-1/2+2 = 5/6 - 3/6 + 12/6 = 14/6 = 7/3

e,3/5+6/11+7/13+2/5+16/11+19/13 = ( 3/5+ 2/5) + ( 6/11 + 16/11) + ( 7/13 + 19/13) = 1 + 2 + 2 = 5

g,75/100+18/21+29/32+1/4+3/21+13/32 = 3/4 + 6/7 + 29/32 + 1/4 + 1/7 + 13/32=  ( 3/4 + 1/4) + ( 29/32 + 13/32) + ( 6/7 + 1/7)= 1 + 21/16 + 1 = 53/16

c: \(=\dfrac{3}{2}\cdot1-1-20=\dfrac{3}{2}-21=\dfrac{-39}{2}\)

\(\frac{3}{1}+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+...+100}\)

\(=3\left(\frac{1}{\frac{1\cdot2}{2}}+\frac{1}{\frac{2\cdot3}{2}}+\frac{1}{\frac{3\cdot4}{2}}+...+\frac{1}{\frac{100\cdot101}{2}}\right)\)

\(=3\left(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+...+\frac{2}{100\cdot101}\right)\)

\(=6\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{100\cdot101}\right)\)

\(=6\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\right)\)

\(=6\left(1-\frac{1}{101}\right)=6-\frac{6}{101}=\frac{606-6}{101}=\frac{600}{101}\)

Ta có:
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
\(...\)
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)     \(\left(1\right)\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
Đặt \(A=\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{3}-\dfrac{1}{100}\)\(< \dfrac{1}{3}\)     \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)

C =\(\frac{1}{100}-\frac{1}{100.99}-...\)\(-\frac{1}{3.2}-\frac{1}{2.1}\)

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

C = 1

mình ko biết dấu sao lag gì nên lam mò nhé

giả sử sao la dấu nhân

suy ra s<1/1.2+1/2.3+...+1/99.100

s<1/1-1/2+1/2-1/3+...+1/99-1/100

s<1/1-1/100

s<99/100<1

suy ra s<1

nếu sao là dấu cộng

suy ra s=+2/2.3+...+2/100.101

1/2s=1/2-1/3+1/3-1/4+...+1/100-1/101

1/2s=1/2-1/100<1/2

1/2 s <1/2 suy ra s<1
 

thanks ban nhiu nha

Ta có:

B=1/2-1/2^2-1/2^3-...-1/2^100

B/2=1/2^2-1/2^3-1/2^4-....-1/2^101

B/2-B=1/2^101-1/2

=>B=(1/2^101-1/2).2

Vậy:B=(1/2^101-1/2).2

Bạn nào làm đc nhanh mình sẽ k

Bạn làm nhanh và đúng nhất mình sẽ k

2A=1-1/2+1/2^2-...+1/2^98-1/2^99

=>3A=1-1/2^100

=>\(A=\dfrac{2^{100}-1}{3\cdot2^{100}}\)