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\(A=1+\frac{1}{2}+...+\frac{1}{16}\)
= \(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{12}\right)+\left(\frac{1}{13}+...+\frac{1}{16}\right)\)
> \(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+4\times\frac{1}{8}+4\times\frac{1}{12}+4\times\frac{1}{16}\)
=\(1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
=\(1+2\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)
= \(1+2\times\frac{13}{12}\)
= \(1+\frac{13}{6}\)
= \(1+2+\frac{1}{6}\)
= \(3+\frac{1}{6}\)>\(3\)
=> \(A>3+\frac{1}{6}>3\)
=> \(A>3+\frac{1}{6}>B\)
=> \(A>B\)
1.
a) \(\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+\frac{6}{99}+\frac{6}{143}\)
\(=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+\frac{6}{9.11}+\frac{6}{11.13}\)
\(=\frac{6}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{6}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{6}{2}.\frac{10}{39}\)
\(=\frac{10}{13}\)
b) \(\frac{3}{24}+\frac{3}{48}+\frac{3}{80}+\frac{3}{120}+\frac{3}{168}\)
\(=\frac{3}{4.6}+\frac{3}{6.8}+\frac{3}{8.10}+\frac{3}{10.12}+\frac{3}{12.14}\)
\(=\frac{3}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{1}{12}-\frac{1}{14}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{4}-\frac{1}{14}\right)\)
\(=\frac{3}{2}.\frac{5}{28}\)
\(=\frac{15}{56}\)
\(a.\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{11.13}\)
\(=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=3.\frac{10}{39}\)
\(=\frac{10}{13}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}>1\)
Ta có\(A=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}\right)\)\(>1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+4\times\frac{1}{8}+4\times\frac{1}{12}+4\times\frac{1}{16}\)
\(=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(=1+2\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)
\(>1+2\times\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\right)=1+2=3=B\)
\(\Rightarrow A>B\)
a , tổng các phân số đã cho là : 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 = 79/64
b, \(\frac{79}{64}\)và \(\frac{2017}{2018}\)= \(\frac{159422}{129152}\)và \(\frac{129088}{129152}\)= \(\frac{159422}{129152}\)> \(\frac{129088}{129152}\)
=> \(\frac{79}{64}\)> \(\frac{2017}{2018}\)
a) 1/2 + 1/4 + 1/8 + 1/ 16 + 1/32 + 1/64
=32/64 + 16/64 + 8/64 + 4/64 + 2/64
=32+16+8+4+2/64 = 66/64= 33/32
b) ta có 33/32 > 1 và 2017/2018<1
nên 33/32 > 2017/2018
a, 2006 x 2004 - \(\frac{2}{1995}\) + 2004 x 2005 = 8038043,999
b, 2006 x 125 + \(\frac{1000}{126}\) x 2006 - 1006 = 265664,6349
c, A = 1991 x 1999
=> A = ( 1995 - 4 ) x ( 1995 + 4 )
A = 1995 x ( 1995 + 4 ) - 4 x ( 1995 + 4 )
A = 1995 x 1995 + 1995 x 4 - ( 4 x 1995 + 4 x 4 )
A = 1995 x 1995 - 4 x 4
mà B = 1995 x 1995
Vậy A < B
d, Gọi giá trị biểu thức là C
C = \(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
C x 2 = \(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}\)
C x 2 = \(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}\)
Vậy C x 2 - C = \(\left(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\right)\)
C = \(\frac{2}{3}-\frac{1}{96}\) ( vì phân số nào có ở số bị trừ cũng có ở số trừ thì trừ hết rồi nên không còn )
C = \(\frac{21}{32}\)
A=1991x1999=(1995-4)1999=1995x1999-4x1999
B=1995x1995=1995x(1999-4)=1995x1999-1995x4>1995x1999-4x1999=A
vậy A<B
Vì \(\frac{1}{33}>\frac{1}{34}>\frac{1}{35}>\frac{1}{36}\)
\(\Rightarrow M>\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\)\(\)
\(\Rightarrow M>\frac{4}{36}=\frac{1}{9}\)
Mà \(\frac{1}{9}>\frac{1}{10}\)
\(\Rightarrow\)\(M>\frac{1}{9}>\frac{1}{10}\)
Vậy : M > N