\(5^{200}=\left(5^2\right)^{100}=25^{100}\)
\(2^{500}=\left(2^5\right)^{100}=32^{100}\)
Vậy \(5^{200}< 2^{500}\)
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\(5^{200}=\left(5^2\right)^{100}=25^{100}\)
\(2^{500}=\left(2^5\right)^{100}=32^{100}\)
Vậy \(5^{200}< 2^{500}\)
1.Tim cac can bac hai ko am cua cac so ko am sau:
a,25
b,2500
c,-52
d.0,49
e,121
Tìm x nguyên biết a)|x+1|+|x+2|+|x+3|+...+|x+100|=2500
b)|x-1|+|x-2|+|x-3|+...+|x-100|=2500
5^x+3 - 5^x+2 =2500
X ( X+1) = 2+4+6+ ...+ 2500
X ( X+1) = 2+4+6+ ...+ 2500
CMR: \(1+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{2500}}< 100\)
Tìm x nguyên biết ; |x-1|+|x-2|+.....+|x-100|=2500
M= \(\dfrac{5}{4}+\dfrac{10}{9}+\dfrac{17}{16}+...+\dfrac{2499}{2500}\)
cho A=3/4+8/9+15/36+..+2499/2500. tính A
Timx:\(\left|x-1\right|+\left|x-2\right|+...+\left|x-100\right|=2500\)