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26 tháng 7 2017
\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{81}\right).\left(1-\frac{1}{100}\right)\)
\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{80}{81}.\frac{99}{100}\)
\(B=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{8.10}{9.9}.\frac{9.11}{10.10}\)
\(B=\frac{1.2.3...8.9}{2.3.4...9.10}.\frac{3.4.5...10.11}{2.3.4...9.10}\)
\(B=\frac{1}{10}.\frac{11}{2}\)
\(B=\frac{11}{20}>\frac{11}{21}\)
23 tháng 8 2018
Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{18}{19}.\frac{19}{20}\)
\(=\frac{1.2....18.19}{2.3...19.20}\)
\(=\frac{1}{20}>\frac{1}{21}\)
Vậy A > 1/21
DD
10 tháng 10 2017
Ta có :
\(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{99}{100}=\frac{3.8.15.....99}{4.9.16.....100}=\frac{1.3.2.4.3.5.....9.11}{2.2.3.3.4.4.....10.10}\)\(=\frac{1.2.3...9}{2.3...10}.\frac{3.4...11}{2.3...10}=\frac{1}{10}.\frac{11}{2}=\frac{11}{20}< \frac{11}{19}\)
10 tháng 10 2017
ta có M = (1- 1/4) (1- 1/9)... ( 1- 1/100)
= 3/2^2.8/3^2 ... 99/10^2
= 1.3/2^2 . 2.4/3^2 ... 9.11/10^ 2
= 1.2.3...9/ 2.3.4...10 . 3.4.5... 11/ 2.3.4... 10
= 1/10 . 11/2 = 11/20 < 11/19
Vậy M < 11/19
NT
1
AM
30 tháng 6 2015
\(\left(\frac{27}{64}\right)^{15}=\frac{\left(3^3\right)^{15}}{\left(2^6\right)^{15}}=\frac{3^{45}}{2^{90}}=\left(\frac{3}{2^2}\right)^{45}\)
\(\left(\frac{81}{256}\right)^{10}=\frac{\left(3^4\right)^{10}}{\left(2^8\right)^{10}}=\frac{3^{40}}{2^{80}}=\left(\frac{3}{2^2}\right)^{40}\)
Do \(\left(\frac{3}{2^2}\right)^{45}
25 tháng 8 2017
a)\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)
\(x=\left(\frac{3}{7}\right)^7\div\left(\frac{3}{7}\right)^5\)
\(x=\left(\frac{3}{7}\right)^2\)
\(x=\frac{9}{49}\)
Vậy...
b)\(\left(-\frac{1}{3}\right)^3.x=\left(\frac{1}{3}\right)^4\)
\(\left(-\frac{1}{3}\right)^3.x=\left(-\frac{1}{3}\right)^4\)
\(x=\left(-\frac{1}{3}\right)^4\div\left(\frac{-1}{3}\right)^3\)
\(x=-\frac{1}{3}\)
Vậy...
c)\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
=>\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
Vậy...
d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)
=>\(x+\frac{1}{4}=\frac{2}{3}\)
\(x=\frac{2}{3}-\frac{1}{4}\)
\(x=\frac{5}{12}\)
Vậy...
Phù, mãi mới xong, tk cho mk nha bn
\(\left(\frac{1}{27}\right)^{23}=\frac{1^{23}}{27^{23}}=\frac{1}{\left(3^3\right)^{23}}=\frac{1}{3^{69}}\)
\(\left(\frac{1}{81}\right)^{16}=\frac{1^{16}}{81^{16}}=\frac{1}{\left(3^4\right)^{16}}=\frac{1}{3^{64}}\)
Vì 369 > 364
\(\frac{1}{3^{69}}< \frac{1}{3^{64}}\)
\(\left(\frac{1}{27}\right)^{23}=\frac{1^{23}}{27^{23}}=\frac{1}{\left(3^3\right)^{23}}=\frac{1}{3^{69}}\)
\(\left(\frac{1}{81}\right)^{16}=\frac{1^{16}}{81^{16}}=\frac{1}{\left(3^4\right)^{16}}=\frac{1}{3^{64}}\)
Vì 369 > 364
=> \(\frac{1}{3^{69}}< \frac{1}{3^{64}}\)
=> \(\left(\frac{1}{27}\right)^{23}< \left(\frac{1}{81}\right)^{16}\)