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\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
\(3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3B-B=1-\dfrac{1}{3^{100}}\)
\(\Rightarrow2B=1-\dfrac{1}{3^{100}}\)
\(0< \dfrac{1}{3^{100}}< 1\Rightarrow0< 1-\dfrac{1}{3^{100}}< 1\)
\(\Rightarrow0< 2B< 1\Rightarrow0< B< \dfrac{1}{2}\Rightarrow\) B không phải số nguyên
Lời giải:
Đặt $\frac{x}{2018}=\frac{y}{2019}=\frac{z}{2020}=a$
$\Rightarrow x=2018a; y=2019a; z=2020a$
$\Rightarrow (x-z)^3=(2018a-2020a)^3=(-2a)^3=-8a^3(1)$
Mặt khác:
$8(x-y)^2(y-z)=8(2018a-2019a)^2(2019a-2020a)=8a^2.(-a)=-8a^3(2)$
Từ $(1); (2)$ ta có đpcm.
\(\dfrac{x}{2018}=\dfrac{y}{2019}=\dfrac{x-y}{-1};\dfrac{y}{2019}=\dfrac{z}{2020}=\dfrac{y-z}{-1};\dfrac{x}{2018}=\dfrac{z}{2020}=\dfrac{x-z}{-2}\\ \Leftrightarrow\dfrac{x-y}{-1}=\dfrac{y-z}{-1}=\dfrac{x-z}{-2}\\ \Leftrightarrow2\left(x-y\right)=2\left(y-z\right)=x-z\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)
\(A=\dfrac{5}{9}.\dfrac{22}{-3}+\dfrac{5}{9}.\dfrac{5}{-3}=\dfrac{5}{9}\left(\dfrac{22}{-3}+\dfrac{5}{-3}\right)=\dfrac{5}{9}.\dfrac{22+5}{-3}=\dfrac{5}{9}.\dfrac{27}{-3}\)
\(A=\dfrac{5.27}{9.-3}=\dfrac{5.27}{-27}=-5\)
\(B=\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(B=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)\(B=\left(6-5-3\right)+\left(\dfrac{-2}{3}-\dfrac{5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
\(B=-2+\left(\dfrac{-2-5+7}{2}\right)+\left(\dfrac{1+3-5}{2}\right)\)
\(B=-2+\dfrac{0}{2}+\left(\dfrac{-1}{2}\right)=-2+0-\dfrac{1}{2}=\dfrac{-5}{2}\)
\(\dfrac{2}{3}A=\dfrac{2}{3}-\left(\dfrac{2}{3}\right)^2+\left(\dfrac{2}{3}\right)^3-...+\left(\dfrac{2}{3}\right)^{2019}-\left(\dfrac{2}{3}\right)^{2020}\)
=>\(\dfrac{5}{3}A=1-\left(\dfrac{2}{3}\right)^{2020}=1-\dfrac{2^{2020}}{3^{2020}}=\dfrac{3^{2020}-2^{2020}}{3^{2020}}\)
=>\(A=\dfrac{3^{2020}-2^{2020}}{3^{2020}}:\dfrac{5}{3}=\dfrac{3^{2020}-2^{2020}}{5\cdot3^{2019}}\) ko là số nguyên