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a) 8180 < 2790
b) 377 > 738
c) 536 < 1124
d) 291 < 535
Đúng thì k, sai thì thôi
\(77^{33}=\left(11.7\right)^{33}=11^{33}.7^{33}=11^{33}.\left(7^3\right)^{11}=11^{33}.343\left(1\right)\)
\(33^{77}=\left(3.11\right)^{77}=3^{77}.11^{77}=\left(3^7\right)^{11}.11^{77}=2187^{11}.11^{77}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow33^{77}>77^{33}\)
a, 97/583 < 13/77
b, \(-\left(\frac{9^{100}+4}{9^{100}-2}\right)< -\left(\frac{9^{100}}{9^{100}-6}\right)\)
chúc bạn hk tốt!!(nhớ k cho mình nha!!@@)
1) Ta có: \(80-\left(5.4^2-4.2^3\right)\)
\(=80-\left(80-32\right)\)
\(=80-80+32\)
\(=32\)
2) \(60-\left[120-\left(42-33\right)^2\right]\)
\(=60-\left(120-9^2\right)\)
\(=60-120+81\)
\(=21\)
\(a,80-\left(5.4^2-4.2^3\right)\)
\(=80-\left(80-32\right)\)
\(=80-80+32\)
\(=32\)
\(b,60-\left[120-\left(42-33\right)^2\right]\)
\(=60-\left(120-9^2\right)\)
\(=60-39\)
\(=21\)
a, \(A=2^0+2^1+2^2+...+2^{2010}\)
\(=>2A=2^1+2^2+2^3+...+2^{2011}\)
\(=>2A-A=\left(2^1+2^2+2^3+...+2^{2011}\right)-\left(2^0+2^1+2^2+...+2^{2010}\right)\)
\(=>2A=2^{2011}-2^0=2^{2011}-1\)
Vì \(2^{2011}-1=2^{2011}-1\)
\(=>A=B\)
a) Ta có : A=1+2+22+...+22010
2A=2+22+23+...+22011
\(\Rightarrow\) 2A-A=(2+22+23+...+22011)-(1+2+22+...+22010)
\(\Rightarrow\) A=22011-1
Mà B=22011-1
\(\Rightarrow\)A=B
Vậy A=B.
b) Ta có : A=2009.2011
B=20102=2010.2010
\(\Rightarrow\)A=2009.2010+2009
B=2009.2010+2010
Vì 2009<2010 nên 2009.2010+2009<2009.2010+2010
hay A<B
Vậy A<B.
Ta có :
\(3^{2n}=\left(3^2\right)^n=9^n\)
\(2^{3n}=\left(2^3\right)^n=8^n\)
Do 9 > 8 => \(9^n>8^n\Rightarrow3^{2n}>2^{3n}\)
Vậy \(3^{2n}>2^{3n}\)
Ta có:
\(3^{2n}=\left(3^2\right)^n=9^n\)
\(2^{3n}=\left(2^3\right)^n=8^n\)
Vì \(9^n>8^n\Rightarrow3^{2n}>2^{3n}\)
_Hok tốt_
!!!