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LT
5
K
5
7 tháng 3 2017
A = 1 + 2 + 22 + ...... + 29
2A = 2(1 + 2 + 22 + ...... + 29)
= 2 + 22 + 23 + ....... + 210
2A - A = (2 + 22 + 23 + ....... + 210) - (1 + 2 + 22 + ...... + 29)
A = 210 - 1
B = 5.28 = (22 + 1).28 = 22.28 + 1.28 = 210 + 28 > 210 - 1
Do đó B > A
ML
7 tháng 3 2017
A = 1 + 2 + 22 + 23 + ... + 29
\(\Rightarrow\)2A = 2 + 22 + 23 + ... + 210
\(\Rightarrow\)2A - A = ( 2 + 22 + 23 + ... + 210 ) - ( 1 + 2 + 22 + 23 + ... + 29 )
\(\Rightarrow\)A = 2 + 22 + 23 + ... + 210 - 1 - 2 - 22 - 23 - ... - 29
A = 210 - 1
Ta có : ( 4 + 1 ).28 = 4.28 + 28 = 28.28 + 28 = 210 + 28
\(\Rightarrow\)210 - 1 < 210 + 28 hay
A > B .
7 tháng 9 2021
\(1,\\ a,2< 3\Rightarrow2^{30}< 3^{30}\Rightarrow-2^{30}>-3^{30}\\ b,6^{10}=6^{2\cdot5}=\left(6^2\right)^5=36^5>35^5\left(36>35\right)\)
\(2,\\ a,\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot5^5\cdot3^5}{5^6\cdot3^{14}}=\dfrac{3}{5}\\ b,\left(8x-1\right)^{2x+1}=5^{2x+1}\\ \Leftrightarrow8x-1=5\\ \Leftrightarrow x=\dfrac{3}{4}\)
7 tháng 9 2021
Bài 2:
a: Ta có: \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)
\(=\dfrac{-3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)
\(=-\dfrac{3}{5}\)
b: Ta có: \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)
\(\Leftrightarrow8x-1=5\)
\(\Leftrightarrow8x=6\)
hay \(x=\dfrac{3}{4}\)
8 tháng 10 2023
a: \(A=\dfrac{3^3\cdot2^3+3^3\cdot2^2+3^3\cdot1}{-13}=\dfrac{27\left(2^3+2^2+1\right)}{-13}=-27\)
b: \(B=\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^3\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)
\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9}{2^{10}\cdot3^7+2^{10}\cdot5\cdot3^8}\)
\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\right)}{2^{10}\cdot3^7\left(1+5\cdot3\right)}=\dfrac{2}{3}\cdot\dfrac{4+27}{1+15}=\dfrac{2}{3}\cdot\dfrac{31}{16}=\dfrac{31}{24}\)
c: \(C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{35}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot2^6-7\right)}=\dfrac{10-9}{5\cdot64-7}=\dfrac{1}{313}\)
9 tháng 10 2021
a, Ta có: \(\left(\dfrac{1}{2}\right)^{300}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)
\(\left(\dfrac{1}{3}\right)^{200}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)
=> \(\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}\)=> \(\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)
b, Ta có: \(\left(\dfrac{1}{3}\right)^{75}=\left[\left(\dfrac{1}{3}\right)^3\right]^{25}=\left(\dfrac{1}{27}\right)^{25}\)
\(\left(\dfrac{1}{5}\right)^{50}=\left[\left(\dfrac{1}{5}\right)^2\right]^{25}\)\(=\left(\dfrac{1}{25}\right)^{25}\)
Do \(\left(\dfrac{1}{27}\right)^{25}< \left(\dfrac{1}{25}\right)^{25}=>\left(\dfrac{1}{3}\right)^{75}< \left(\dfrac{1}{5}\right)^{50}\)
Kiểm tra lại bài nhé, học tốt!!
DM
1
SY
1
29 tháng 8 2020
a) 8.(-5).(-4).2 = 8.20.2 = 8.40 = 320
b) \(1\frac{3}{7}+\left(-\frac{1}{3}+2\frac{4}{7}\right)\)
\(=1\frac{3}{7}-\frac{1}{3}+2\frac{4}{7}\)
\(=\left(1\frac{3}{7}+2\frac{4}{7}\right)-\frac{1}{3}=\left(1+2\right)+\left(\frac{3}{7}+\frac{4}{7}\right)-\frac{1}{3}=4-\frac{1}{3}=\frac{11}{3}\)
c) \(\frac{8}{5}\cdot\frac{-2}{3}+\frac{-5\cdot5}{3\cdot5}\)
\(=\frac{8}{5}\cdot\frac{-2}{3}+\frac{-25}{15}=\frac{-16}{15}+\frac{-25}{15}=\frac{-41}{15}\)
d) \(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}\left(-2\right)^2=\frac{6}{7}+\frac{5}{8}\cdot\frac{1}{5}-\frac{3}{16}\cdot4\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{4}=\frac{13}{56}\)
8 tháng 4 2018
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(A< 1-\frac{1}{9}=\frac{8}{9}=B\)
\(\Rightarrow\)\(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
A=1+2+2^2+2^3+....+2^9
2A=2+2^2+2^3+....+2^10
2A-A=2^10-1
A=2^10-1/2
B=5.2^8=(2^2+1).2^8=2^10+2^8
=>B>A
2A = 2(1 + 2 + 22 + .... + 29 )
= 2 + 22 + 23 + ..... + 210
2A - A = (2 + 22 + 23 + ..... + 210) - (1 + 2 + 22 + .... + 29 )
A = 210 - 1
B = 5.28 = (22 + 1).28 = 210 + 28
210 - 1 < 210 + 28
=> A < B