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\(\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{2}{2016}+\dfrac{1}{2017}\)
\(=\left(\dfrac{2016}{2}+1\right)+\left(\dfrac{2015}{3}+1\right)+...+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{1}{2017}+1\right)+1\)
\(=\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\dfrac{2018}{2018}\)
\(=2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)
Theo đề, ta có: \(x=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}}=2018\)
\(\dfrac{1}{3}B=\dfrac{1}{3^2}+\dfrac{1}{3^3}.....\dfrac{1}{3^{2018}}\)
\(B-\dfrac{1}{3}B=\left(\dfrac{1}{3}+\dfrac{1}{3^2}...\dfrac{1}{3^{2017}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}...\dfrac{1}{3^{2018}}\right)\)
\(\dfrac{2}{3}B=\dfrac{1}{3}-\dfrac{1}{3^{2018}}\)
\(B=\dfrac{1}{2}-\dfrac{3}{2}.\dfrac{1}{3^{2018}}\) <\(\dfrac{1}{2}\)
\(S=\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2017}{2^{2016}}\)
\(\Rightarrow2S=2+\dfrac{3}{2^1}+\dfrac{4}{2^2}+...+\dfrac{2017}{2^{2015}}\)
\(\Rightarrow2S-S=\left(2+\dfrac{3}{2^1}+\dfrac{4}{2^2}+...+\dfrac{2017}{2^{2015}}\right)-\left(\dfrac{2}{2^1}+\dfrac{3}{2^2}+...+\dfrac{2017}{2^{2016}}\right)\)
\(\Leftrightarrow S=2+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}-\dfrac{2017}{2^{2016}}\)
Tới đây thì đơn giản rồi nhé
Theo bài ra, ta có: \(B=\dfrac{2018}{1}+\dfrac{2017}{2}+\dfrac{2016}{3}+...+\dfrac{1}{2018}\)
\(B=\left(\dfrac{2018}{1}+1\right)+\left(\dfrac{2017}{2}+1\right)+\left(\dfrac{2016}{3}+1\right)+...+\left(\dfrac{1}{2018}+1\right)-2018\)
\(B=2019+\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}-2018\)
\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\left(2019-2018\right)\)
\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+1\)
\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\dfrac{2019}{2019}\)
\(B=2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)\)
Khi đó:\(\dfrac{B}{A}=\dfrac{2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}}\)
\(\Rightarrow\dfrac{B}{A}=2019\), là 1 số nguyên.
Vậy \(\dfrac{B}{A}\) là số nguyên.
\(a)3\dfrac{1}{2}.\dfrac{4}{49}-\left[2,\left(4\right):2\dfrac{5}{11}\right]:\left(\dfrac{-42}{5}\right)\)
\(=\dfrac{7}{2}.\dfrac{4}{49}-\dfrac{88}{27}:\left(\dfrac{-42}{7}\right)\)
\(=\dfrac{2}{7}-\dfrac{-220}{567}\)
\(=\dfrac{382}{567}\)
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Chúc bạn học tốt!
a: \(A=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{-5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\)
\(=\dfrac{-3}{5}+\dfrac{3}{5}=0\)
b: \(=3^4-\left(-8\right)^2-\left(-25\right)^2\)
\(=81-64-625=-608\)
c: \(=2^3+3\cdot1\cdot\dfrac{1}{4}\cdot4+\left[4:\dfrac{1}{2}\right]:8\)
\(=8+3+4\cdot2:8=11+1=12\)
\(A=\left(\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}\right)+\left(\dfrac{-6}{13}+\dfrac{1}{2}+1\dfrac{1}{3}\right)\)
\(A=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}-\dfrac{6}{13}+\dfrac{1}{2}+\dfrac{4}{3}\)
\(A=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{7}{13}+\dfrac{6}{13}\right)+\left(\dfrac{4}{3}-\dfrac{1}{3}\right)\)
\(A=1-1+1=1\)
\(B=\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right).\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}\)
\(B=\dfrac{-3}{2}:\dfrac{3}{-4}.\dfrac{-9}{2}-\dfrac{1}{4}\)
\(B=2.\dfrac{-9}{2}-\dfrac{1}{4}\)
\(=-9-\dfrac{1}{4}=\dfrac{-37}{4}\)
\(a,A=\left(\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}\right)+\left(-\dfrac{6}{13}+\dfrac{1}{2}+1\dfrac{1}{3}\right)\)
\(A=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}+\dfrac{-6}{13}+\dfrac{1}{2}+\dfrac{4}{3}\)
\(A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{7}{13}-\dfrac{6}{13}\right)+\left(-\dfrac{1}{3}+\dfrac{4}{3}\right)\)
\(A=-1+1=0\)
\(b,B=\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right)\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}\)
\(B=\left(-\dfrac{3}{2}.\dfrac{-4}{3}\right).\dfrac{-9}{2}-\dfrac{1}{4}\)
\(B=8.\dfrac{-9}{2}-\dfrac{1}{4}\)
\(B=-36-\dfrac{1}{4}\)
B = \(-\dfrac{145}{4}\)
b)\(\frac{1}{9}.\frac{2}{145}-4\frac{1}{3}.\frac{2}{145}+\frac{2}{145}\)
\(=\frac{2}{145}.\left(\frac{1}{9}-\frac{13}{3}+1\right)\)
\(=\frac{2}{145}.\left(-\frac{29}{9}\right)\)
\(=\frac{-2}{45}\)
Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
...
\(\dfrac{1}{2017^2}< \dfrac{1}{2016.2017}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2017^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2016.2017}\)\(\Rightarrow A< 1-\dfrac{1}{2017}< 1\)
\(\Rightarrow[A]=0\)