Cho hỏi 3(2^2+1)(2^4+1)....(2^64+1)+1 tính làm sao
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1/2+1/4+1/8+1/16+1/32+1/64
=2x(1/2+1/4+1/8+1/16+1/321/64)
=1+1/2+1/4+1/8+1/16+1/32
=1+1/2+1/4+1/8+1/16+1/32-1/2-1/4-1/8-1/16-1/32-1/64
=1-1/64
=63/64
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(=\frac{32}{64}+\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{1}{64}+\frac{1}{64}\)
\(=\frac{62}{64}=\frac{31}{32}\)
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Ý bạn là mỗi phép tính phải thêm dấu và ....để kết quả bằng 6 phải không ạ
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A = \(\dfrac{1}{2}\)+ \(\dfrac{1}{3}\)+ \(\dfrac{1}{4}\)+ \(\dfrac{1}{2}\)+ \(\dfrac{2}{3}\)+ \(\dfrac{3}{4}\)
A = ( \(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\)) + ( \(\dfrac{1}{4}\)+\(\dfrac{3}{4}\)) +( \(\dfrac{1}{3}\)+ \(\dfrac{2}{3}\))
A = 1 + 1 + 1
A = 3
B = ( \(\dfrac{3}{4}\) + \(\dfrac{5}{7}\)+ \(\dfrac{16}{64}\)+ \(\dfrac{2}{7}\)) - 2
B = ( \(\dfrac{3}{4}\)+ \(\dfrac{1}{4}\)) + ( \(\dfrac{5}{7}\)+ \(\dfrac{2}{7}\)) - 2
B = 1 + 1 - 2
B = 2 - 2
B = 0
tính nhanh hộ mình câu này nha (có kèm bài làm , ai nhanh mình tick cho );
1/2+1/4+1/8+1/16+1/32+1/64
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(1/2+1/8)+(1/4+1/16)+(1/8+1/32)+1/64
=1/10+1/20+1/40+1/64
=61/320
lấy 1/64 làm mẫu xố chung
1/64+2/64+4/64+8/64+12/64+32/64=59/64