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291 và 535
291 = (213)7 = 81927
535 = (55)7 = 31257
Vì 81927 > 31257 => 291 > 535
Vậy 291 > 535
\(B=\frac{31}{2}.\frac{32}{2}.....\frac{60}{2}\)
\(B=\left(31.32.33....60\right).\frac{1.2.3....60}{2^{30.\left(1.2.3...30\right)}}\)
\(B=\left(1.3.5.....59\right).\frac{2.4.6.....60}{2.4.6....60}=1.3.5...59\)
=> \(B=A\)
a ) 53 54 < 96 97 . b ) 93 102 > 23 32 . c ) − 299 300 < − 101 102 . d ) − 163 167 > − 223 227
\(2023A=\dfrac{2023^{31}+4046}{2023^{31}+2}=1+\dfrac{4044}{2023^{31}+2}\)
\(2023B=\dfrac{2023^{32}+4046}{2023^{32}+2}=1+\dfrac{4044}{2023^{32}+2}\)
mà 2023^31+2<2023^32+2
nên A>B
A=32019+1+3+32+33+...+32018
⇒A=1+3+32+...+32018+32019
⇒3A=3×(1+3+3^2+3^3+....+3^2019)
3A=3+3^2+3^3+....+3^2020
3A-A=(3+3^2+3^3+....+3^2020) -(1+3+3^2+....+3^2019)
2A= 3^2020-1
⇒ A =( 3^2020-1):2
A=32019+1+3+32+33+...+32018
⇒A=1+3+32+...+32018+32019
⇒3A=3×(1+3+3^2+3^3+....+3^2019)
⇒3A=3+3^2+3^3+....+3^2020
⇒3A-A=(3+3^2+3^3+....+3^2020) -(1+3+3^2+....+3^2019)
⇒2A= 3^2020-1
⇒ A =( 3^2020-1):2
\(3B=3+3^2+3^3+...+3^{2019}\\ 2B=3^{2019}-1\\ B=\dfrac{3^{2019}-1}{2}\)
\(9B=3^2+3^4+...+3^{2020}\)
\(\Leftrightarrow8B=3^{2018}-1\)
\(\Leftrightarrow B=\dfrac{3^{2018}-1}{8}\)
Bài 2:
\(A=3+3^2+...+3^{2018}\)
\(\Leftrightarrow3A=3^2+3^3+...+3^{2019}\)
\(\Leftrightarrow2A=3^{2019}-3\)
hay \(A=\dfrac{3^{2019}-3}{2}=\dfrac{3^{2019}+9-12}{2}=\dfrac{3\left(3^{2018}+3\right)-12}{2}\)
=>A>B