A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
Giúp mình nha, mính sẽ tick lại cho
TÍNH NHANH:\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
GIẢI CHI TIẾT GIÚP MÌNH AI NHANH MÌNH TICK CHO
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
A=14 +18 +116 +132 +164 +1128 +1256 +1512
=12 −14 +14 −18 +....+1256 −1512
=12 −1512
=255512
Vậy A=255512
Phạm Long Khánh
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
GIẢI CHI TIẾT GIÚP MÌNH NHÉ MÌNH CẢM ƠN
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
=1/2-1/4+1/4-1/8+1/8-....+1/156-1/152
=1/2-1/152
=255/512
A=255/512
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(A=\frac{2-1}{4}+\frac{2-1}{8}+\frac{2-1}{16}+\frac{2-1}{32}+\frac{2-1}{64}+\frac{2-1}{128}+\frac{2-1}{256}+\frac{2-1}{512}\)
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{256}-\frac{1}{512}\)
\(A=\frac{1}{2}-\frac{1}{512}\)
\(A=\frac{256}{512}-\frac{1}{512}=\frac{255}{512}\)
\(y=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
Tính y
Chỉ có nhửng bạn có lời giải hay mình mới tick đúng
2y=\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
2y-y=1-\(\frac{1}{256}\)
y=\(\frac{255}{256}\)
1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256
Quy đồng (MSC: 256). Ta có:
= 128/256+64/256+32/256+16/256+8/256+4/256+2/256+1/256
= 128+64+32+16+8+4+2+1/256
= 255/256
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
Nhận xét :
1/2 = 1 - 1/2 ; 1/4 = 1/2 - 1/4 ; 1/8 = 1/4 - 1/8 ; ..... ; 1/256 = 1/128 - 1/256
=> A = 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + ..... + 1/128 - 1/256
=> A = 1 - 1/256 = 255/256
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{156}\)
Ta có:
\(\frac{1}{2}=1-\frac{1}{2}\) \(;\) \(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)\(;\) \(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)\(;...;\) \(\frac{1}{256}=\frac{1}{128}-\frac{1}{256}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)
\(\Rightarrow A=1-\frac{1}{256}\)
\(\Rightarrow A=\frac{256}{256}-\frac{1}{256}\)
\(\Rightarrow A=\frac{255}{256}\)
Vậy \(A=\frac{255}{256}\)
◘_◘ Đúng 100%
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
quy đồngcác phân số lấy mẫu số là 512 .ta có tử số là
256 +128 + 64 +32 +16 +8 +4 +2 +1 =495
A =\(\frac{495}{512}\)
cho hỏi làm thế nào để nó ra phân số như thế kia zạ
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
Ta có:
2A=1+1/2+1/4+1/8+1/16+1/32+1/64
=> 2A-A=(1+1/2+1/4+1/8+1/16+1/32+1/64) - (1/2+1/4+1/8+1/16+1/32+1/64+1/128)
A= 1-1/128=127/128
\(\frac{1}{2}+\frac{1}{4}=1-\frac{1}{4}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=1-\frac{1}{8}\)
............................................
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}=1-\frac{1}{128}=\frac{127}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\)
= 1 - 1/2+ 1/2- 1/4 +1/4 - 1/8 +1/8 -1/16 +1/16 -1/32 +1/32 -1/64 +1/64 - 1/128
= 1-1/128
=127/128
\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)= \(\frac{64}{128}\)+ \(\frac{32}{128}\)+ \(\frac{16}{128}\)+ \(\frac{8}{128}\)+ \(\frac{4}{128}\)+ \(\frac{2}{128}\)+ \(\frac{1}{128}\).
= \(\frac{127}{128}\).
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)
= \(1\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{8}\)+ \(\frac{1}{8}\)- \(\frac{1}{16}\)+ \(\frac{1}{16}\)- \(\frac{1}{32}\)+ \(\frac{1}{32}\)- \(\frac{1}{64}\)+ \(\frac{1}{64}\)- \(\frac{1}{128}\)
= \(1\)- \(\frac{1}{128}\)
= \(\frac{127}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}\)
\(=1-\frac{1}{128}\)
\(\frac{127}{128}\)
Đặt dãy trên là A
\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)
\(\Leftrightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\right)\)
\(\Leftrightarrow A=1-\frac{1}{128}+0+0+...+0\)
\(\Leftrightarrow A=\frac{127}{128}\)