Tính Nhanh:
B=\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)
Giúp mình với mình cần gấp ạ
Bài 1: Tính nhanh
C = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\)
Em xin hỏi cách giải bài này, em cảm ơn ạ.
C = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\)
2\(\times\)C = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)
2 \(\times\) C - C = 1 - \(\dfrac{1}{128}\)
C = \(\dfrac{127}{128}\)
\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{128}\)+\(\dfrac{1}{256}\)
Tính nhanh
Tính :
a) A= \(\dfrac{1}{2010.2009}-\dfrac{1}{2009.2008}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
b) B= 63.(-124)-124.(-37)+(\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\)).(-66)
Giúp mình với ạ! Mình cần gấp lắm
\(a,A=\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2009}+\dfrac{1}{2008}-...-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\\ A=1+\dfrac{1}{2010}=\dfrac{2011}{2010}\)
\(b,B=\left(-124\right)\left(63-37\right)+\dfrac{17}{66}\left(-66\right)=-124\cdot26+17=-3224+17=-3207\)
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 = `\(\dfrac{341}{1024}\)
Tính nhanh:
\(\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)
c.7m28dm2=.....m2
\(\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{64}-\dfrac{1}{128}\\ =\dfrac{1}{2}-\dfrac{1}{128}\\ =\dfrac{63}{128}\)
\(7m^28dm^2=7,08m^2\)
Đặt \(A=\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{128}\)
\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^7}\\ \Rightarrow2A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^6}\\ \Rightarrow2A-A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^6}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^7}\\ \Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2^7}=\dfrac{2^6-1}{2^7}=\dfrac{63}{128}\)
\(7m^28dm^2=7\dfrac{8}{100}m^2=7,08m^2\)
A = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\)+ \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\) + \(\dfrac{1}{256}\)
\(A=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{128}-\dfrac{1}{256}\right)\)
\(A=1-\dfrac{1}{256}\)
\(A=\dfrac{255}{256}\)
\(\dfrac{1}{27}+a^3\\ 8x^3+27y^3\\ \dfrac{1}{8}x^3+8y^3\\ x^6+1\\ x^9+1\\ x^3-64\\ x^3-125\\ 8x^6-27y^3\\ \dfrac{1}{64}x^6-125y^3\\ \dfrac{1}{8}x^3-8\\ x^3+6x^2+12x+8\\ x^3+9x^2+27x+27\) Giúp mình với mình cần gấp ;-;
1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)
2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)
4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)
6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)
7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)
8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)
10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)
11) \(=\left(x+2\right)^3\)
12) \(=\left(x+3\right)^3\)
S=\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{128}\)+\(\dfrac{1}{256}\)+\(\dfrac{1}{512}\)+\(\dfrac{1}{1024}\)
Chứng minh rằng: \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^3}\)+\(\dfrac{1}{4^2}+....+\dfrac{1}{100^2}< \dfrac{3}{4}\)
Giúp mình với.Mình cần gấp ạ