Question

[nguyễn việt hà]

Tính:

a. M=[ (-4)³ + 4³ ] : ( 1 + 3 + 5 + ... + 2005 )

b. N=1-2-3+4+5-6-7+8+...+2001+2005

Answer 1

[H o o n i e - )]

a)M=[(−4)3+43]:(1+3+5+...+2005)

M=\left[-64+64\right]\cdot(1+3+5+...+2005)M=[−64+64]⋅(1+3+5+...+2005)

M=0\cdot(1+3+5+...+2005)M=0⋅(1+3+5+...+2005)

M=0M=0

b, Như câu a

Answer 2

[Huỳnh Quang Sang]

\(a)M=\left[(-4)^3+4^3\right]:(1+3+5+...+2005)\)

\(M=\left[-64+64\right]\cdot(1+3+5+...+2005)\)

\(M=0\cdot(1+3+5+...+2005)\)

\(M=0\)

b, Tương tự