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2431:11=221 5083:221= 23 => 5083/2431=5083:221/2431:221=23/11
2431:11=221
5083:221= 23
=> 5083/2431=5083:221/2431:221=23/11
`291/891=(97xx3)/(297xx3)=97/297`
`549/1281=(183xx3)/(183xx7)=3/7`
`3672/4284=(6xx612)/(7xx612)=6/7`
`7696/9970=(2xx3848)/(2xx4985)=3848/4985`
`578/3757=(2xx289)/(13xx289)=2/13`
`5083/2431=(23xx221)/(11xx221)=23/11`
\(\dfrac{291}{891}=\dfrac{97}{297};\dfrac{549}{1281}=\dfrac{3}{7}\\ \dfrac{3672}{4284}=\dfrac{6}{7};\dfrac{7976}{9970}=\dfrac{4}{5}\\ \dfrac{578}{3757}=\dfrac{2}{13};\dfrac{5083}{2431}=\dfrac{23}{11}\)
a , tổng các phân số đã cho là : 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 = 79/64
b, \(\frac{79}{64}\)và \(\frac{2017}{2018}\)= \(\frac{159422}{129152}\)và \(\frac{129088}{129152}\)= \(\frac{159422}{129152}\)> \(\frac{129088}{129152}\)
=> \(\frac{79}{64}\)> \(\frac{2017}{2018}\)
a) 1/2 + 1/4 + 1/8 + 1/ 16 + 1/32 + 1/64
=32/64 + 16/64 + 8/64 + 4/64 + 2/64
=32+16+8+4+2/64 = 66/64= 33/32
b) ta có 33/32 > 1 và 2017/2018<1
nên 33/32 > 2017/2018
\(2A=1+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{18}+\dfrac{1}{32}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)
\(A=1-\dfrac{1}{64}\)
\(A=\dfrac{63}{64}\)
giúp em với ạ , em cấn gấp , 4h em học rồi ạ . Cảm ơn những ai giúp em ạ , em tick luôn ạ
tính nhanh p/s 1+ 5/4 + 5/8 + 5/16 + 5/32 + 5/64
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
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
\(A=1-\frac{1}{64}\)
\(A=\frac{63}{64}\)
=32/64+14/64+8/64+4/64+2/64+1/64
=32+14+8+4+2+1/64
=61/64
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)
\(=1-\frac{1}{64}\)
\(=\frac{63}{64}\)
\(\frac{12}{15}\times\frac{22}{47}+\frac{25}{47}:\frac{15}{12}\)
\(\frac{12}{15}\times\frac{22}{47}+\frac{25}{47}\times\frac{12}{15}\)
\(=\frac{12}{15}\times\left(\frac{22}{47}+\frac{25}{47}\right)\)
\(=\frac{12}{15}\times\frac{22+25}{47}\)
\(=\frac{12}{15}\times1=\frac{12}{15}\)
Hoặc
Cách phía trên là cách lớp 5 nhé
Ta có:\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(=\dfrac{32}{64}+\dfrac{16}{64}+\dfrac{8}{64}+\dfrac{4}{64}+\dfrac{2}{64}+\dfrac{1}{64}=\dfrac{63}{64}\)
Ta có:\(\dfrac{5083}{2431}=\dfrac{23}{11}\)