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\(\frac{3}{4}+\frac{2}{3}-\frac{1}{6}\)
\(=\frac{3}{4}+\frac{2}{3}\)
\(=\frac{17}{12}\)
\(=\frac{17}{12}-\frac{1}{6}\)
\(=\frac{90}{72}\)
Gọi tổng trên là A
Ta có : \(A=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+...+\frac{1}{2560}\)
\(2A=2\left(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{2560}\right)\)
\(2A=\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+...+\frac{1}{1280}\)
\(2A-A=\left(\frac{2}{5}+\frac{1}{5}+...+\frac{1}{1280}\right)-\left(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{2560}\right)\)
\(A\left(2-1\right)=\frac{2}{5}-\frac{1}{2560}\)
\(A.1=\frac{1024}{2560}-\frac{1}{2560}\)
\(A=\frac{1023}{2560}\)
Ta có : A = 1/5 + 1/10 + 1/20 + ... + 1/2560
2A = 2 ( 1/5 + 1/10 + ... + 1/2560 )
2A = 2/5 + 1/5 + 1/10 + .. + 1/2560
2A - A = ( 2/5 + 1/5 + ... + 1/1280 ) - ( 1/5 + 1/10 + ... + 1/2560 )
A = 2 - 1 = 2/5 - 1/2560
A.1 = 1024/2560 - 1/2560
A = 1023 = 2560
\(a,\left(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right):\frac{1}{4}:\frac{1}{6}\)
\(=\left(\frac{1}{6}+\frac{1}{4}-\frac{1}{5}\right)\cdot\frac{1}{4}\cdot\frac{1}{6}\)
\(=\left(\frac{10}{60}+\frac{15}{60}-\frac{12}{60}\right)\cdot\frac{1}{24}\)
\(=\frac{13}{60}\cdot\frac{1}{24}\)
\(=\frac{13}{1440}\)
\(b,\frac{2006\cdot2005-1}{2004\cdot2006+2005}\)
\(\frac{2006\cdot2005-1}{2004\cdot2006+2005}\)
\(=\frac{2006\cdot\left(2004+1\right)-1}{2004 \cdot2006+2005}\)
\(=\frac{2006\cdot2004+2006\cdot1-1}{2004\cdot2006+2005}\)
\(=\frac{2006\cdot2004+2005}{2004\cdot2006 +2005}=1\)
Mình nghĩ phần b, ko có cách 2 đâu bạn .
\(A>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)
\(A>\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2015-2014}{2014.2015}\)
\(A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(A>1-\frac{1}{2015}\)
Mà \(\frac{1}{2015}< \frac{1}{4}\Rightarrow1-\frac{1}{2015}>1-\frac{1}{4}=\frac{3}{4}\Rightarrow A>\frac{3}{4}\)
a,=25/6;7/6=25/6x6/7=25/7
b,=7/2x32/6=56/3
c,=17/5-11/10=34/10-11/10=23/10
d,=8/3+11/4=32/12+33/12=65/12
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
=\(\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
=\(\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{4}-\frac{3}{8}+\frac{3}{8}-\frac{3}{16}+\frac{3}{16}-\frac{3}{36}\)
=\(\frac{3}{1}-\frac{3}{36}\)=\(\frac{35}{12}\)
10/12-1/2-1/3
=10/12-6/12-4/12
=4/12-4/12
=0
10/12-1/2-1/3=1/3-1/3
=0