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\(=\dfrac{2}{2}\).(\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+...+\(\dfrac{2}{x.\left(x+1\right)}\))
=2.(\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+...+\(\dfrac{2}{x.\left(x+1\right)}\))
=2.(\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+...+\(\dfrac{1}{x.\left(x+1\right)}\))
=2.[(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\))+(\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\))+(\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\))+...+(\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\))
=2.[\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\)]
2.[(\(\dfrac{1}{3}\)-\(\dfrac{1}{3}\))+(\(\dfrac{1}{4}\)-\(\dfrac{1}{4}\))+...+(\(\dfrac{1}{x}\)-\(\dfrac{1}{x}\))+(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))]
=2.[0+0+...+0+(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))]
=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))
=2.(\(\dfrac{1.x+1-1.2}{2.x+1}\))
=2.(\(\dfrac{x+1-2}{2x}\))=2.\(\dfrac{x-1}{2x}\)=\(\dfrac{2.\left(x-1\right)}{2x}\)=\(\dfrac{2x-2}{2x}\)
\(\dfrac{2x-2}{2x}\)=\(\dfrac{2014}{2016}\)\(\Rightarrow\)(2x-2).2016=2014.2x=4032x-4032=4028x
\(\Rightarrow\)4032x-4028x=4x=4032\(\Rightarrow\)x=4032:4=1008
Đặt A=\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x.\left(x+1\right)}\)
\(A=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}\)
\(A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x.\left(x+1\right)}\)
3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
\(\dfrac{2}{3^2}+\dfrac{2}{4^2}+\dfrac{2}{5^2}+....\dfrac{2}{2016^2}\)
Ta thấy: \(\dfrac{2}{3^2}< \dfrac{2}{2.3}\)
\(\dfrac{2}{4^2}< \dfrac{2}{3.4}\)
...\(\dfrac{2}{2016^2}< \dfrac{2}{2015.2016}\)
Đặt:A=\(\dfrac{2}{3^2}+\dfrac{2}{4^2}+\dfrac{2}{5^2}+...+\dfrac{2}{2016^2}\)
=>\(A< \dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{2015.2016}\)
=>\(A< \dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+\dfrac{2}{4}-\dfrac{2}{5}+...+\dfrac{2}{2015}-\dfrac{2}{2016}\)
=>A<\(\dfrac{2}{2}-\dfrac{2}{2016}\)
=>A<\(\dfrac{1007}{1008}\) mà \(\dfrac{1007}{1008}\) < 1
=>A<1
Vậy \(\dfrac{2}{3^2}+\dfrac{2}{4^2}+\dfrac{2}{5^2}+...+\dfrac{2}{2016^2}\)<1 (\(đpcm\))
\(\dfrac{2}{3^2}+\dfrac{2}{4^2}+...+\dfrac{2}{2016^2}=2\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\right)\)
Ta có: \(\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{2016^2}< \dfrac{1}{2015.2016}\)
\(\Rightarrow2\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\right)< 2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}\right)\)
\(\Rightarrow2\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\right)< 2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)
\(\Rightarrow2\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\right)< 2\left(\dfrac{1}{2}-\dfrac{1}{2017}\right)=1-\dfrac{2}{2017}< 1\)
=> đpcm
a) \(\dfrac{1}{6},\dfrac{1}{3},\dfrac{1}{2},\dfrac{2}{3}\)
Vì \(\dfrac{1}{6}:\dfrac{1}{2}=\dfrac{1}{3}\) ; \(\dfrac{1}{3}:\dfrac{2}{3}=\dfrac{1}{2}\) nên suy ra \(\dfrac{1}{2}:\dfrac{3}{4}=\dfrac{2}{3}\)
Vậy số để điền vào chỗ trống là \(\dfrac{2}{3}\).
b) \(\dfrac{1}{8},\dfrac{5}{24},\dfrac{7}{24},\dfrac{3}{8}\)
Vì \(\dfrac{1}{8}:\dfrac{3}{5}=\dfrac{5}{24}\) ; \(\dfrac{5}{24}:\dfrac{5}{7}=\dfrac{7}{24}\) nên suy ra \(\dfrac{7}{24}:\dfrac{7}{9}=\dfrac{3}{8}\)
Vậy số để điền vào chỗ trống là \(\dfrac{3}{8}\).
c) \(\dfrac{1}{5},\dfrac{1}{4},\dfrac{3}{10},\dfrac{7}{20}\)
Vì \(\dfrac{1}{5}:\dfrac{4}{5}=\dfrac{1}{4}\) ; \(\dfrac{1}{4}:\dfrac{5}{6}=\dfrac{3}{10}\) nên suy ra \(\dfrac{3}{10}:\dfrac{6}{7}=\dfrac{7}{20}\)
Vậy số để điền vào chỗ trống là \(\dfrac{7}{20}\).
a) Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in Z\right)\)
\(\Rightarrow B=\dfrac{5^{12}+2}{5^{13}+2}< 1\)
\(B< \dfrac{5^{12}+2+48}{5^{13}+2+48}\Rightarrow B< \dfrac{5^{12}+50}{5^{13}+50}\Rightarrow B< \dfrac{5^2\left(5^{10}+2\right)}{5^2\left(5^{11}+2\right)}\Rightarrow B< \dfrac{5^{10}+2}{5^{11}+2}=A\)\(B< A\)
bạn ơi thế còn phần b thì sao? Mong bạn có câu trả lời sớm tớ cảm ơn bạn nhiều lắm
a) \(S=\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}\)
\(\Rightarrow2S=\dfrac{2\left(2+2^2+2^3+...+2^{2017}\right)}{1-2^{2017}}\)
\(2S=\dfrac{2^2+2^3+2^4+...+2^{2018}}{1-2^{2017}}\)
\(\Rightarrow2S-S=S=\dfrac{2^2+2^3+2^4+...+2^{2018}}{1-2^{2017}}-\dfrac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}\)
\(S=\dfrac{\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)}{1-2^{2017}}\)
\(S=\dfrac{2^{2018}-2}{1-2^{2017}}=\dfrac{-2\left(1-2^{2017}\right)}{1-2^{2017}}=-2\) vậy \(S=-2\)
1. Ta có: \(\dfrac{a}{b}>1\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\left(m\in Z\right)\)
\(B=\dfrac{2016^{2016}}{2016^{2016}-3}>\dfrac{2016^{2016}+2}{2016^{2016}-3+2}=\dfrac{2016^{2016}+2}{2016^{2016}-1}=A\)
Vậy A > B
2. \(\dfrac{1}{2016.2015}+\dfrac{1}{2015.2014}+\dfrac{1}{2014.2013}+...+\dfrac{1}{1.2}\)
= \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\)
= \(1-\dfrac{1}{2016}\)
=\(\dfrac{2015}{2016}\)