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Bài 1: 1/3+1/9+1/27+1/81+1/243+1/729
Đặt:
A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
Nhân A với 3 ta có:
\(Ax3=3+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow Ax3-S=3-\frac{1}{243}\)
\(\Rightarrow2A=\frac{2186}{729}\)
\(\Rightarrow A=\frac{2186}{729}:2\)
\(\Rightarrow A=\frac{1093}{729}\)
a) = \(\frac{127}{96}\)
b) = \(\frac{255}{256}\)
c) Mik bỏ nha
d) = \(\frac{1023}{512}\)
e) = \(\frac{2343}{625}\)
b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
a ) \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\times1280\)
= \(\frac{1}{2}\times1280+\frac{1}{4}\times1280+\frac{1}{8}\times1280+\frac{1}{16}\times1280+\frac{1}{32}\times1280+\frac{1}{64}\times1280\)\(+\frac{1}{128}\times1280\)
= 640 + 320 + 160 + 80 + 40 + 20 + 10
= ( 640 + 160 ) + ( 320 + 80 ) + ( 40 + 20 + 10 )
= 800 + 400 + 70
= 1270
a, 68 + 42 x 5 - 625 : 25
= 68 + 210 - 25
= 278 - 25
= 253
b, 15 x { 37 + 8 + 20 } + 15 x { 13 + 22 }
= 15 x 65 + 15 x 35
= 15 x { 65 + 35 }
= 15 x 100
= 1500
c, 125 x 56 x 3
= 7000 x 3
= 21000
d, 249 - { 49 - 100 }
= 249 - 49 + 100
= 200 + 100
= 300
e, 1/2 + 1/4 + 1/ 8 + 1/32 + 1/64 + 1/128
= 1/1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128
= 1/1 - 1/128
= 127/128
A, \(68+42\times5-625:25\)
= 68 +210 - 25
= 278 - 25
= 253
B,\(15\times\left(37+8+20\right)+15\times\left(13+22\right)\)
= \(15\times\left(37+8+20+13+22\right)\)
=\(15\times100\)
= 1500
C,\(125\times56\times3\)
= \(125\times8\times7\times3\)
= \(1000\times21\)
= 21000
D, 249 - { 49 - 100 }
= 249 - 49 -100
= 200 - 100
= 100
E, \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)(mình sửa lại đề chút xíu nha, nó cứ bị sai sai sao ấy)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}\)
= \(1-\frac{1}{128}\)
=\(\frac{128}{128}-\frac{1}{128}\)
= \(\frac{127}{128}\)
TRA LOI:
1+5/4+5/8+5/16+5/32+5/64=219/64
3+3/5+3/25+3/125+3/512=3,8