Question

[Quỳnh Như]

1, Tính

a, 2008 . 2008 - 2010 . 2006

b, \(\dfrac{232323.29}{23.292929}\)

c, \(\dfrac{\left(2^{17}+5^{17}\right).\left(3^{14}-5^{12}\right).\left(2^4-4^2\right)}{15^2+5^3+67^7}\)

2, So sánh \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\) với 1

Answer 1

[Nguyễn Lê Phước Thịnh]

Câu1:

a: \(=2008^2-\left(2008-2\right)\left(2008+2\right)\)

\(=2008^2-\left(2008^2-4\right)\)

=4

b: \(=\dfrac{23\cdot29\cdot10101}{23\cdot29\cdot10101}=1\)

c: \(=\dfrac{\left(2^{17}+5^{17}\right)\left(3^{14}-5^{12}\right)\cdot\left(16-16\right)}{15^2+5^3+67^7}\)

=0