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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 .
TT
10 tháng 8 2020
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
\(\Leftrightarrow A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)....\left(\frac{1}{4052169}-1\right)\left(\frac{1}{\text{}\text{}4056196}-1\right)\)
\(\Leftrightarrow A=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....\frac{-4056195}{\text{}4056196}\)
\(\Leftrightarrow A=\frac{\left(-1\right)3}{2^2}.\frac{\left(-2\right)4}{3^3}.\frac{\left(-3\right)5}{4^2}.....\frac{\left(-2013\right)2015}{\text{}2014^2}\)
\(\Leftrightarrow A=\frac{\left(-1\right)\left(-2\right)....\left(-2013\right)}{2.3...1014}.\frac{3.4......2015}{2.3......2014}\)
\(\Leftrightarrow A=\frac{-1}{1014}.\frac{2015}{2}=\frac{-2015}{4028}\)
VÌ \(\frac{-2015}{4028}< \frac{-1}{2}\)
\(\Rightarrow A< \frac{-1}{2}\Leftrightarrow A< B\)
XO
10 tháng 8 2020
Ta có \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2014^2}-1\right)=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-4056195}{2014^2}\)
\(=-\left(\frac{1.3}{2^2}.\frac{2.4}{3^2}...\frac{2013.2015}{2014^2}\right)=-\left(\frac{1.3.2.4...2013.2015}{2.2.3.3...2014.2014}\right)\)
\(=-\left(\frac{\left(1.2.3...2013\right)\left(3.4.5...2015\right)}{\left(2.3.4...2014\right)\left(2.3.4...2014\right)}\right)=-\frac{2015}{2014.2}=-\frac{2015}{4028}< \frac{-2014}{4028}< \frac{1}{2}=B\)
=> A < B
20 tháng 9 2019
\(a,\left[2^{17}+16^2\right]\cdot\left[9^{15}-3^{15}\right]\cdot\left[2^4-4^2\right]\)
\(=\left[2^{17}+16^2\right]\cdot\left[9^{15}-3^{15}\right]\cdot\left[16-16\right]\)
\(=\left[2^{17}+16^2\right]\left[9^{15}-3^{15}\right]\cdot0=0\)
\(b,\left[8^{2017}-8^{2015}\right]\cdot\left[8^{2014}\cdot8\right]\)
\(=8^{2015}\left[8^2-1\right]\cdot8^{2015}\)
\(=8^{2015}\cdot63\cdot8^{2015}=8^{4030}\cdot63\)sửa lại câu b , có vấn đề rồi
\(c,\frac{2^8+8^3}{2^5\cdot2^3}=\frac{2^8+\left[2^3\right]^3}{2^5\cdot2^3}=\frac{2^8+2^9}{2^8}=\frac{2^8\left[1+2\right]}{2^8}=3\)
2.a, \(2^6=\left[2^3\right]^2=8^2\)
Mà 8 = 8 nên 82 = 82 hay 26 = 82
b, \(5^3=5\cdot5\cdot5=125\)
\(3^5=3\cdot3\cdot3\cdot3\cdot3=243\)
Mà 125 < 243 nên 53 < 35
c, 26 = [23 ]2 = 82
Mà 8 > 6 nên 82 > 62 hay 26 > 62
d, 7200 = [72 ]100 = 49100
6300 = \(\left[6^3\right]^{100}\)= 216100
Mà 49 < 216 nên 49100 < 216100 hay 7200 < 6300
NT
1
EC
3 tháng 9 2016
Câu 1:
a) 2225 và 3150
Ta có:2225=(29)25=51225
3150=(36)25=72925
Vì 51225<72925
Suy ra: 2225<3150
EC
3 tháng 9 2016
Câu 2:
a)\(25^3:5^2=\left(5^2\right)^3:5^2=5^6:5^2=5^4\)
b)\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
c)\(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3+\frac{1}{4}:2=3+\frac{1}{8}=\frac{25}{8}\)
Câu 3:
a)\(9.3^3.\frac{1}{81}.3^2=3^2.3^3.3^2.\left(\frac{1}{3^4}\right)=3^7:3^4=3^3\)
b)\(4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:\left(2^3.\frac{1}{2^4}\right)=2^7:\frac{1}{2}=2^8\)
c)\(3^2.2^5.\left(\frac{2}{3}\right)^2=288.\frac{4}{9}=2^7\)
d)\(\left(\frac{1}{3}\right)^3.\frac{1}{3}.9^2=\left(\frac{1}{3}\right)^4.\left(3^2\right)^2=3^4.\left(\frac{1}{3}\right)^4=3^4:3^4=1\)
QK
0
4 tháng 7 2017
1.\(45^{10}.5^{30}=45^{10}.125^{10}=\left(45.125\right)^{10}=5625^{10}\)
2.a. \(\left(2x-1\right)^3=-8\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)
b.\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}\)
c. \(\left(2x+3\right)^2=\frac{9}{121}\Leftrightarrow\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{cases}}\)
d.\(\left(3x-1\right)^3=-\frac{8}{27}=\left(-\frac{2}{3}\right)^3\)
\(\Leftrightarrow3x-1=-\frac{2}{3}\Leftrightarrow x=\frac{1}{9}\)
4.
a.\(99^{20}=\left(99^2\right)^{10}=9801^{10}\)
Do \(9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)
b.\(3^{4000}=\left(3^2\right)^{2000}=9^{2000}\)
\(\Rightarrow3^{4000}=9^{2000}\)
c.\(2^{332}=\left(2^3\right)^{110}.2^2=8^{110}.4\)
\(3^{223}=\left(3^2\right)^{110}.3^3=\left(3^2\right)^{110}.9=9^{110}.9\)
Ta thấy \(4.8^{110}< 9.9^{110}\)
Vậy \(2^{332}< 3^{223}\)
A=210-1<210+28=22.28+1.28=(22+1).28=5.28=B
=>A<B
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