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a) \(\frac{3}{4} + \left( {\frac{1}{2} - \frac{1}{3}} \right) = \frac{9}{{12}} + \left( {\frac{6}{{12}} - \frac{4}{{12}}} \right) = \frac{9}{{12}} + \frac{2}{{12}} = \frac{{11}}{{12}}\)
\(\frac{3}{4} + \frac{1}{2} - \frac{1}{3} = \frac{9}{{12}} + \frac{6}{{12}} - \frac{4}{{12}} = \frac{{15}}{{12}} - \frac{4}{{12}} = \frac{{11}}{{12}}\)
Vậy \(\frac{3}{4} + \left( {\frac{1}{2} - \frac{1}{3}} \right)\) = \(\frac{3}{4} + \frac{1}{2} - \frac{1}{3}\)
b)\(\frac{2}{3} - \left( {\frac{1}{2} + \frac{1}{3}} \right) = \frac{4}{6} - \left( {\frac{3}{6} + \frac{2}{6}} \right) = \frac{4}{6} - \frac{5}{6} = \frac{{ - 1}}{6}\)
\(\frac{2}{3} - \frac{1}{2} - \frac{1}{3} = \frac{4}{6} - \frac{3}{6} - \frac{2}{6} = \frac{1}{6} - \frac{2}{6} = \frac{{ - 1}}{6}\)
Vậy \(\frac{2}{3} - \left( {\frac{1}{2} + \frac{1}{3}} \right)\)=\(\frac{2}{3} - \frac{1}{2} - \frac{1}{3}\).
`#3107`
`a)`
`3/4 + (1/2 - 1/3)`
`= 3/4 + (3/6 - 2/6)`
`= 3/4 + 1/6`
`= 11/12`
`3/4 + 1/2 - 1/3`
`= 9/12 + 6/12 - 4/12`
`= (9 + 6 - 4)/12`
`= 11/12`
Vì `11/12 = 11/12`
`=> 3/4 + (1/2 - 1/3) = 3/4 + 1/2 - 1/3`
`b)`
`2/3 - (1/2 + 1/3)`
`= 2/3 - (3/6 + 2/6)`
`= 2/3 - 5/6`
`= -1/6`
`2/3 - 1/2 - 1/3`
`= 4/6 - 3/6 - 2/6`
`= (4 - 3 - 2)/6`
`= -1/6`
Vì `-1/6 = -1/6`
`=> 2/3 - (1/2 + 1/3) = 2/3 - 1/2 - 1/3`
Ta có : \(\frac{1}{n^2}-1=\frac{1-n^2}{n^2}=\frac{\left(1-n\right)\left(1+1\right)}{n^2}\)
Áp dụng :
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)
\(=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.\frac{-3.5}{4.4}.....\frac{-2013.2015}{2014.2014}\)
\(=\frac{-\left(1.2.3...2013\right)\left(3.4.5....2015\right)}{\left(2.3.4.....2014\right)\left(2.3.4......2014\right)}=\frac{-2015}{2014.2}=\frac{-2015}{4028}\)
Sr còn thiếu
\(A=-\frac{2015}{4028}< \frac{-2014}{4028}=-\frac{1}{2}\)
Vậy \(A< B\)
Ta có:
\(2^2<4^2\Rightarrow\frac{1}{2^2}>\frac{1}{4^2}\)
\(3^2<6^2\Rightarrow\frac{1}{3^2}>\frac{1}{6^2}\)
\(4^2<8^2\Rightarrow\frac{1}{4^2}<\frac{1}{8^2}\)
\(...\)
\(100^2<200^2\Rightarrow\frac{1}{100^2}>\frac{1}{200^2}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}\)
\(\Rightarrow A>B\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)\cdot\cdot\cdot\cdot\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
\(A=\left(\frac{-3}{4}\right)\left(\frac{-8}{9}\right)\left(\frac{-15}{16}\right)\cdot\cdot\cdot\left(\frac{-4052168}{4052169}\right)\left(\frac{-4056195}{4056196}\right)\)
\(A=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot\frac{-3\cdot5}{4\cdot4}\cdot....\cdot\frac{-2012\cdot2014}{2013\cdot2013}\cdot\frac{-2013\cdot2015}{2014\cdot2014}\)
\(A=\frac{-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot....\cdot\left(-2012\right)\cdot\left(-2013\right)}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\cdot\frac{3\cdot4\cdot5\cdot....\cdot2014\cdot2015}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\)
\(A=\frac{-1}{2014}\cdot\frac{2015}{2}=\frac{-2015}{4028}\)
Ta thấy \(\frac{-2015}{4028}< \frac{-1}{2}\) \(\Rightarrow A< B\)
