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26 tháng 3 2023

Ta có C = \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}\)

2C = 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{19}}\)

2C - C = ( 1 + \(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{19}}\) ) - ( \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}\) )

C = 1 - \(\dfrac{1}{2^{20}}=\dfrac{2^{20}-1}{2^{20}}\)

a: x+2/5=1/2

=>x=1/2-2/5=5/10-4/10=1/10

b; x-2/5=2/7

=>x=2/7+2/5=10/35+14/35=24/35

c: 3/5-x=1/10

=>x=3/5-1/10=6/10-1/10=5/10=1/2

d: x*3/4=9/20

=>x=9/20:3/4=9/20*4/3=36/60=3/5

e: x:1/7=14

=>x=14*1/7=2

f: =>x+1/4=2/5:1/2=4/5

=>x=4/5-1/4=16/20-5/20=11/20

g: =>x*2/3=9/12+2/3=3/4+2/3=9/12+8/12=17/12

=>x=17/12:2/3=17/12*3/2=51/24=17/8

Giải

Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)

\(\Rightarrow\)D < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)

Nhận xét: \(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{19.20}=\dfrac{1}{19}-\dfrac{1}{20}\)

\(\Rightarrow\) D< 1- \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

D< 1 - \(\dfrac{1}{20}\)

D< \(\dfrac{19}{20}\)<1

\(\Rightarrow\)D< 1

Vậy D=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{5^2}\)<1

30 tháng 4 2017

A=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

A=\(\dfrac{1}{2^2.1}+\dfrac{1}{2^2.2^2}+\dfrac{1}{3^2.2^2}+...+\dfrac{1}{50^2.2^2}\)

A=\(\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)

\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\right)\)

Ta có :

\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\dfrac{1}{50.50}< \dfrac{1}{49.50}\)

\(\Rightarrow A< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)Nhận xét :

\(\dfrac{1}{1.2}< 1-\dfrac{1}{2};\dfrac{1}{2.3}< \dfrac{1}{2}-\dfrac{1}{3};...;\dfrac{1}{49.50}< \dfrac{1}{49}-\dfrac{1}{50}\)

\(\Rightarrow A< \dfrac{1}{2^2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

A<\(\dfrac{1}{2^2}\left(1-\dfrac{1}{50}\right)\)

A<\(\dfrac{1}{4}.\dfrac{49}{50}\)<1

A<\(\dfrac{49}{200}< \dfrac{1}{2}\)

\(\Rightarrow A< \dfrac{1}{2}\)

a: =11+3/4-6-5/6+4+1/2+1+2/3

=10+9/12-10/12+6/12+8/12

=10+13/12=133/12

b: \(=2+\dfrac{17}{20}-1-\dfrac{11}{15}+2+\dfrac{3}{20}\)

=3-11/15

=34/15

c: \(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)

\(=\dfrac{31}{7}:\dfrac{31}{5}=\dfrac{5}{7}\)

d: \(=\dfrac{29}{8}\cdot\dfrac{36}{29}\cdot\dfrac{15}{23}\cdot\dfrac{23}{5}=\dfrac{9}{2}\cdot3=\dfrac{27}{2}\)

14 tháng 4 2017

a) \(\left(\dfrac{3}{20}-\dfrac{1}{5}x\right)\cdot1\dfrac{2}{3}=1\dfrac{1}{4}\)

\(\left(\dfrac{3}{20}-\dfrac{1}{5}x\right)\cdot\dfrac{5}{3}=\dfrac{5}{4}\)

\(\dfrac{3}{20}-\dfrac{1}{5}x=\dfrac{5}{4}:\dfrac{5}{3}\\ \dfrac{3}{20}-\dfrac{1}{5}x=\dfrac{3}{4}\\ \dfrac{1}{5}x=\dfrac{3}{20}-\dfrac{3}{4}\\ \dfrac{1}{5}x=-\dfrac{3}{5}\\ x=-\dfrac{3}{5}:\dfrac{1}{5}\\ x=-3\)

b) \(\dfrac{-2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)

\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\\ \dfrac{-2}{3}x=\dfrac{1}{10}\\ x=\dfrac{1}{10}:\dfrac{-2}{3}\\ x=-\dfrac{3}{20}\)

c) \(-\dfrac{2}{3}-\dfrac{1}{3}\left(2x-7\right)=\dfrac{3}{2}\)

\(\dfrac{1}{3}\left(2x-7\right)=-\dfrac{2}{3}-\dfrac{3}{2}\\ \dfrac{1}{3}\left(2x-7\right)=-\dfrac{13}{6}\\ 2x-7=-\dfrac{13}{6}:\dfrac{1}{3}\\ 2x-7=-\dfrac{13}{2}\\ 2x=-\dfrac{13}{2}+7\\ 2x=\dfrac{1}{2}\\ x=\dfrac{1}{4}\)

23 tháng 4 2023

1) \(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)

\(=\dfrac{1}{2}+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}+\dfrac{5}{18}\)

\(=\dfrac{3}{2}-\dfrac{4}{9}+\dfrac{5}{18}\)

\(=\dfrac{19}{18}+\dfrac{5}{18}\)

\(=\dfrac{24}{18}\)

\(=\dfrac{4}{3}\)

2) \(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-1+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=\dfrac{1}{15}+\dfrac{7}{15}\)

\(=\dfrac{8}{15}\)

3) \(\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}\)

\(=\left(\dfrac{-11}{31}-\dfrac{20}{31}\right)+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=-1+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=\dfrac{1}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=\dfrac{1}{3}-\dfrac{1}{10}\)

\(=\dfrac{7}{30}\)

4) \(\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)

\(=\dfrac{5}{7}.-\dfrac{7}{11}\)

\(=-\dfrac{35}{77}\)

\(=-\dfrac{5}{11}\)

a:=>0,75x-x+1,25x=0,2

=>x=0,2

b: =>1/3-x=-3/6+4/6=1/6

=>x=1/3-1/6=1/6

c: =>(x-1)/45=-6/30=-1/5

=>x-1=-9

=>x=-8

d: =>(2/5x-1)=-5/7

=>2/5x=2/7

=>x=5/7