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a) Ta có: \(A=1^3+2^3+3^3+...+100^3\)
\(=\left(1-1\right)\cdot1\cdot\left(1+1\right)+1+\left(2-1\right)\cdot2\cdot\left(2+1\right)+2+...+\left(100-1\right)\cdot100\cdot\left(100+1\right)+100\)
\(=1+2+1\cdot2\cdot3+...+99\cdot100\cdot101\)
\(=5050+25497450\)
\(=25502500\)
A = 20 + 21 + 22 + ...... + 2100
=> 2A= 21+...+2101
=>2A-A=A=( 21 + 22 + ...... + 2101)-(20 + 21 + 22 + ...... + 2100)
A=2101-1
cái còn lại tương tự thôi
- Ta co
2A=\(2^1+2^2+2^3+......+2^{101}\)
2A -A= \(2^1+2^2+2^3+.....+2^{101}-2^0-2^1-2^2.......-2^{100}\)
A = \(2^{101}-2^0\)
A = \(2^{101}-1\)
Cac cau con lai tuong tu cau tren.
A=\(2^2-9^3+4^{-2}.16-2.5^2\)
\(=4-729+1-50=-774\)
B=\(\left(2^3.2\right).\dfrac{1}{2}+3^{-2}.3^2-7.1+5\)
\(B=2^4.\dfrac{1}{2}+1-7+5=8+1-7+5=7\)
1. So sánh
a) \(25^{50}\) và \(2^{300}\)
\(25^{50}=25^{1.50}=\left(25^1\right)^{50}=25^{50}\)
\(2^{300}=2^{6.50}=\left(2^6\right)^{50}=64^{50}\)
Vì \(25< 64\) nên \(25^{50}< 64^{50}\)
Vậy \(25^{50}< 2^{300}\)
b) \(625^{15}\) và \(12^{45}\)
\(625^{15}=625^{1.15}=\left(625^1\right)^{15}=625^{15}\)
\(12^{45}=12^{3.15}=\left(12^3\right)^{15}=1728^{15}\)
Vì \(625< 1728\) nên \(625^{15}< 1728^{15}\)
Vậy \(625^{15}< 12^{45}\)
1.So sánh
a)\(25^{50}\) và \(2^{300}\)
Ta có : \(2^{300}=\left(2^6\right)^{50}=64^{50}\)
Vì \(25^{50}< 64^{50}\) nên \(25^{50}< 2^{300}\)
b)\(625^{15}\) và \(12^{45}\)
Ta có : \(12^{45}=\left(12^3\right)^{15}=1728^{15}\)
Vì \(625^{15}< 1728^{15}\) nên \(625^{15}< 12^{45}\)
\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
= \(\frac{1}{4}+\frac{1}{2}\)
= \(\frac{3}{4}\)
b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)
=\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)
= \(-\frac{35}{27}+\frac{47}{21}\)
= \(\frac{178}{189}\)
c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)
= \(\frac{117}{13}-\frac{311}{65}\)
= \(\frac{274}{65}\)
d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)
= \(\frac{1}{3}+\frac{5}{2}\)
= \(\frac{17}{6}\)
\(a,A=2^0+2^1+2^2+....+\)\(2^{2010}\)
\(\Rightarrow2A=2^1+2^2+2^3+....+2^{2011}\)
\(2A-A=\left(2^1+2^2+2^3+...+2^{2011}\right)-\left(2^0+2^1+2^2+...+2^{2010}\right)\)
\(A=2^{2011}-2^0\)
\(A=2^{2011}-1\)
\(b,B=1+3+3^2+...+3^{100}\)
\(\Rightarrow3B=3+3^2+3^3+...+3^{101}\)
\(3B-B=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)
\(2B=3^{101}-1\)
\(\Rightarrow B=\frac{3^{101}-1}{2}\)
\(c,C=4+4^2+4^3+...+4^n\)
\(\Rightarrow4C=4^2+4^3+4^4+...+4^{n+1}\)
\(4C-C=\left(4^2+4^3+4^4+...+4^{n+1}\right)-\left(4+4^2+4^3+...+4^n\right)\)
\(3C=4^{n+1}-4\)
\(\Rightarrow C=\frac{4^{n+1}-4}{3}\)
\(d,D=1+5+5^2+...+5^{2000}\)
\(\Rightarrow5D=5+5^2+5^3+...+5^{2001}\)
\(5D-D=\left(5+5^2+5^3+...+5^{2001}\right)-\left(1+5+5^2+...+5^{2000}\right)\)
\(4D=5^{2001}-1\)
\(\Rightarrow D=\frac{5^{2001}-1}{4}\)
b)
B=1+3+3^2+3^3+..+3^100
=> 3B = 3 + 3^2 + 3^3 + ...+ 3^101
=> 3B - B = ( 3 + 3^2 + 3^3 + ...+ 3^101) - (1+3+3^2+3^3+..+3^100)
=> 2B = 3^101 - 1
=> B =( 3^101 - 1) / 2