\(\dfrac{1}{2\cdot6}\)+\(\dfrac{1}{4\cdot9}\) +\(\dfrac{1}{6\cdot12}\) +....+\(\dfrac{1}{36\cdot57}\) +\(\dfrac{1}{38\cdot60}\) dấu * là dấu nhân . mọi ng cho mình các phép tính cụ thể với ạ.
\(\dfrac{1\cdot2\cdot4+2\cdot4\cdot8+4\cdot8\cdot16+8\cdot16\cdot32}{1\cdot3\cdot4+2\cdot6\cdot8+4\cdot12\cdot16+8\cdot24\cdot32}\). dấu . là nhân .mọi ng cho m các phép tính cụ thể ạ.
\(\dfrac{1.2.4+2.4.8+4.8.16+8.16.32}{1.3.4+2.6.8+4.12.16+8.24.32}\)
\(=\dfrac{8.\left(1+8+4.16+16.32\right)}{12.\left(1+8+4.16+16.32\right)}\)
\(=\dfrac{8}{12}=\dfrac{2}{3}\)
Tính: \(2015-\frac{1}{2\cdot6}-\frac{1}{4\cdot9}-\frac{1}{6\cdot12}-...-\frac{1}{36\cdot57}-\frac{1}{38\cdot60}\)
ai tính hộ tui với
\(\frac{1}{2\cdot6}+\frac{1}{4\cdot9}+\frac{1}{6\cdot12}+...+\frac{1}{36\cdot57}+\frac{1}{38\cdot60}\)
Ta có:\(\frac{1}{2\times6}+\frac{1}{4\times9}+...+\frac{1}{36\times57}+\frac{1}{38\times60}\)
\(=\frac{1}{6}\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{18\times19}+\frac{1}{19\times20}\right)\)
\(=\frac{1}{6}\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=\frac{1}{6}\times\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{6}\times\frac{19}{20}=\frac{19}{120}\)
(đây chắc là toán lớp 7,bạn ạ)
Đặt A= 1/2.6 + 1/4.9 + 1/6.12 + ... + 1/36.57 + 1/38.60
A= 1/2.1.2.3 + 1/2.2.3.3 + 1/2.3.3.4 + ... + 1/2.18.3.19 + 1/2.19.3.20
A= 1/1.2.6 + 1/2.3.6 + 1/3.4.6 + ... + 1/18.19.6 + 1/19.20.6
A= 1/6 . ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/18.19 + 1/19.20)
A= 1/6 . ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/18 - 1/19 + 1/19 - 1/20)
A= 1/6 . ( 1-1/20)
A= 1/6 . 19/20
A= 19/120
Đặt A= 1/2.6 + 1/4.9 + 1/6.12 + ... + 1/36.57 + 1/38.60
A= 1/2.1.2.3 + 1/2.2.3.3 + 1/2.3.3.4 + ... + 1/2.18.3.19 + 1/2.19.3.20
A= 1/1.2.6 + 1/2.3.6 + 1/3.4.6 + ... + 1/18.19.6 + 1/19.20.6
A= 1/6 . ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/18.19 + 1/19.20)
A= 1/6 . ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/18 - 1/19 + 1/19 - 1/20)
A= 1/6 . ( 1-1/20)
A= 1/6 . 19/20
A= 19/120
\(\dfrac{2}{1\cdot3}\)+\(\dfrac{2}{3\cdot5}\) +\(\dfrac{2}{5\cdot7}\)+\(\dfrac{2}{7\cdot9}\)+\(\dfrac{2}{9\cdot11}\) dấu . là nhân nh.mọi ng cho m lời giải chi tiết vs ạ.
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\\ =1-\dfrac{1}{11}=\dfrac{10}{11}\)
\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\\ =1-\dfrac{1}{11}\\ =\dfrac{10}{11}\)
= \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)
\(=1-\dfrac{1}{11}\)
\(=\dfrac{10}{11}\)
M=\(\dfrac{1}{4}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{256}\)+\(\dfrac{1}{1024}\). mọi ng cho mình lời giải cụ thể với ạ.
M = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024
4.M = 1 + 1/4 + 1/16 + 1/64 + 1/256
4M - M = (1 + 1/4 + 1/16 + 1/64 + 1/256 ) - ( 1/4 + 1/16 + 1/64 + 1/256 + 1/1024 )
3M = 1 - 1/1024
3M = 1023/1024
M = 341/1024
M=\(\dfrac{1}{4}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{256}\)+\(\dfrac{1}{1024}\)
=\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\)+\(\dfrac{1}{4^5}\)
=>4M=1+\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\)
=>4M-M=3M=(1+\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\))-(\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\)+\(\dfrac{1}{4^5}\))=1-\(\dfrac{1}{4^5}\)=\(\dfrac{1023}{1024}\)
=>M=\(\dfrac{1023}{1024}\):3=\(\dfrac{341}{1024}\)
`@` `\text {Ans}`
`\downarrow`
\(M=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}+\dfrac{1}{1024}\)
\(\Rightarrow4M=1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}\)
\(\Rightarrow4M-M=\) \(\left(1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}\right)-\left(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}+\dfrac{1}{1024}\right)\)
\(=\) \(1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256}-\dfrac{1}{1024}\)
\(=1-\dfrac{1}{1024}\)
\(=\dfrac{1024}{1024}-\dfrac{1}{1024}=\dfrac{1023}{1024}\)
`4M - M = 3M`
\(\Rightarrow3M=\dfrac{1023}{1024}\)
\(\Rightarrow M=\dfrac{1023}{1024}\div3\)
\(\Rightarrow M=\dfrac{341}{1024}\)
Vậy, `M = `\(\dfrac{341}{1024}\)
Chứng tỏ:
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}=\dfrac{1}{4}\)
(Lưu ý dấu "." trong phân số là dấu nhân)
Giúp mình với các bạn!!!
#Toán lớp 6\(\dfrac{1}{1.2.3}\) + \(\dfrac{1}{2.3.4}\) + .....+ \(\dfrac{1}{10.11.12}\)
= \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}\) + \(\dfrac{1}{2.3}\) - \(\dfrac{1}{3.4}\) +....+ \(\dfrac{1}{10.11}\) - \(\dfrac{1}{11.12}\)
=\(\dfrac{1}{1.2}\) + (- \(\dfrac{1}{2.3}\) + \(\dfrac{1}{2.3}\))+.......+ ( \(-\dfrac{1}{10.11}\) + \(\dfrac{1}{10.11}\)) - \(\dfrac{1}{11.12}\)
=\(\dfrac{1}{2}\) - \(\dfrac{1}{11.12}\) =\(\dfrac{1}{2}\) - \(\dfrac{1}{132}\) =\(\dfrac{66}{132}\)-\(\dfrac{1}{132}\) =\(\dfrac{65}{132}\) Vì \(\dfrac{33}{132}\) = \(\dfrac{1}{4}\) nên \(\dfrac{65}{132}\) > \(\dfrac{1}{4}\)tính nhanh:
\(\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+4\cdot8\cdot12}{1\cdot3\cdot4+4\cdot6\cdot8+6\cdot9\cdot12+8\cdot12\cdot16}\)
dấu \(\cdot\)là dấu nhân nha!
=\(\frac{6\left(1+8+27+64\right)}{12\left(1+16+54+128\right)}\)
=\(\frac{6.100}{12.199}\)
=\(\frac{50}{199}\)
Tk mình với nha mọi người!!!!!
\(\frac{1x2x3+2x4x6+3x6x9+4x8x12}{1x3x4+4x6x8+6x9x12+8x12x16}\)
\(\frac{6x\left(1+8+27+64\right)}{12x\left(1+16+54+128\right)}=\frac{6x100}{12x199}=\frac{50}{199}\)
1/2*6+1/4*9+1/6*12+.....+1/36*57 +1/38*60.dấu * là nhân nha . mọi ng cho mình lời giải cụ thể với ạ
\(\dfrac{1}{3}\)+\(\dfrac{5}{6}\)*(x-\(\dfrac{11}{5}\))=\(\dfrac{3}{4}\).mọi ng cho mình lời giải chi tiết và cụ thể vs ạ. đây là bài tìm x ạ.
\(\dfrac{1}{3}+\dfrac{5}{6}\cdot\left(x-\dfrac{11}{5}\right)=\dfrac{3}{4}\)
\(\dfrac{5}{6}\cdot\left(x-\dfrac{11}{5}\right)=\dfrac{3}{4}-\dfrac{1}{3}\)
\(\dfrac{5}{6}\cdot\left(x-\dfrac{11}{5}\right)=\dfrac{5}{12}\)
\(x-\dfrac{11}{5}=\dfrac{5}{12}\cdot\dfrac{6}{5}\)
\(x-\dfrac{11}{5}=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+\dfrac{11}{5}\)
\(x=\dfrac{27}{10}\)
\(\dfrac{5}{6}\left(x-\dfrac{11}{5}\right)=\dfrac{3}{4}-\dfrac{1}{3}\)
\(\dfrac{5}{6}\left(x-\dfrac{11}{5}\right)=\dfrac{5}{12}\)
\(x-\dfrac{11}{5}=\dfrac{5}{12}:\dfrac{5}{6}\)
\(x-\dfrac{11}{5}=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+\dfrac{11}{5}=\dfrac{27}{10}\)
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{1}{3}+\dfrac{5}{6}\times\left(x-\dfrac{11}{5}\right)=\dfrac{3}{4}\)
\(\dfrac{5}{6}\times\left(x-\dfrac{11}{5}\right)=\dfrac{3}{4}-\dfrac{1}{3}\)
\(\dfrac{5}{6}\times\left(x-\dfrac{11}{5}\right)=\dfrac{5}{12}\)
\(x-\dfrac{11}{5}=\dfrac{5}{12}\div\dfrac{5}{6}\)
\(x-\dfrac{11}{5}=\dfrac{5}{12}\times\dfrac{6}{5}\)
\(x-\dfrac{11}{5}=\dfrac{6}{12}\)
\(x=\dfrac{6}{12}+\dfrac{11}{5}\)
\(x=\dfrac{27}{10}\)