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2021x0.5+2021x1%-2021:2
=2021x0.5+2021x0.01-2021x0.5
=2021x(0.5+0.01-0.5)
=2021x 0.01
=20,21
\(=2021\cdot2\cdot\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)=4042\cdot\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)=0\)
\(...=2021.\left(1000...000-1\right)\) (21 chữ số 0)
\(...=2021.\left(10^{2021}-1\right)\)
Ta có:
\(A=\frac{2021^{2021}+1}{2021^{2022}+1}\Leftrightarrow10A=\frac{2021^{2022}+10}{2021^{2022}+1}=1+\frac{9}{2021^{2022}+1}\)
\(B=\frac{2021^{2022}-1}{2021^{2023}-1}\Leftrightarrow10B=\frac{2021^{2023}-10}{2021^{2023}-1}=1-\frac{9}{2021^{2023}-1}\)
Hay ta đang so sánh: \(\frac{9}{2021^{2022}};\frac{9}{2021^{2023}}\)
Mà \(\frac{9}{2021^{2022}}>\frac{9}{2021^{2023}}\)nên \(\frac{2021^{2021}+1}{2021^{2022}+1}>\frac{2021^{2022}-1}{2021^{2023}-1}\)hay\(A>B\)
Vậy \(A>B\)
Bài 1:
A = \(\dfrac{1}{1\times3}\) + \(\dfrac{1}{3\times5}\) + \(\dfrac{1}{5\times7}\) +...+ \(\dfrac{1}{2019\times2021}\)
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{2}{1\times3}\) + \(\dfrac{2}{3\times5}\) + \(\dfrac{2}{5\times7}\)+...+ \(\dfrac{2}{2019\times2021}\))
A = \(\dfrac{1}{2}\) \(\times\)( \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+...+ \(\dfrac{1}{2019}\) - \(\dfrac{1}{2021}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{2021}\))
A = \(\dfrac{1010}{2021}\)
a, 2021/1 = 2021
b, 2021/10>1
c,2021/2021=1
d,1/2021<1
suy ra : phân số bé nhất là: d,1/2021
bạn lam 1234 kia im mõm đi
mk mới chữa câu này xong =.=