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Ta có : \(\frac{10^{2019}-1}{10^{2018}-1}< \frac{10^{2019}-1+11}{10^{2018}-1+11}=\frac{10^{2019}+10}{10^{2018}+10}=\frac{10\left(10^{2018}+1\right)}{10\left(10^{2017}+1\right)}=\frac{10^{2018}+1}{10^{2017}+1}\)
Vậy \(\frac{10^{2019}-1}{10^{2018}-1}< \frac{10^{2018}+1}{10^{2017}+1}\)
Ta có : \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right)}=\frac{2018}{2019}\)
\(\Rightarrow\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2018}{2019}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2018}{2019}=\frac{1}{2019}\)
\(\Rightarrow x+1=2019\)
\(\Rightarrow x=2018\)
Vậy x = 2018
Nhớ t.i.c.k cho mình nha!
Chỗ \(x(x+1)\Rightarrow\frac{1}{x(x+1)}\) nhé
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x(x+1)}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x(x+1)}=\frac{2018}{2019}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2018}{2019}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2019}\Leftrightarrow x+1=2019\Leftrightarrow x=2018\)
26 x 84 + 74 x 85
= 26 x 84 + 74 x 84 + 74
= 84 x (26 + 74) + 74
= 84 x 100 + 74
= 840 + 74
= 914
\(\frac{2017\times2018+2019}{2019\times2018-2017}\)
= \(\frac{2019\times2018-2\times2018+2019}{2019\times2018-2017}\)
= \(\frac{2019\times2018-4036+2019}{2019\times2018-2017}\)
= \(\frac{2019\times2018-2017}{2019\times2018-2017}\)
= 1
\(\dfrac{2019+2018.2057}{2019.2057-38}\)\(=\dfrac{2018}{-38}\)