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T
18 tháng 6 2019
#)Giải :
a) \(A=\frac{4^5.9^4-2^6.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2^{10}.3^8.3}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^8.3}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=-\frac{1}{3}\)
18 tháng 6 2019
\(a,A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{-1}{3}\)
Học tốt!!!!!!!!!!!!!
19 tháng 4 2017
1) Ta có : \(\frac{x-2}{4}=\frac{5+x}{3}\)
\(\Rightarrow\left(x-2\right).3=\left(5+x\right).4\)
\(\Rightarrow3x-6=20+4x\)
\(\Rightarrow3x=26+4x\)
\(\Rightarrow3x=26+x+3x\)
\(\Rightarrow0=26+x\)
\(\Rightarrow x=0-26\)
\(\Rightarrow x=-26\)
2) Ta có : \(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)
\(\Rightarrow\frac{1}{A}=1+2+2^2+...+2^{2012}\)
\(\Rightarrow\frac{2}{A}=2+2^2+2^3+...+2^{2013}\)
\(\Rightarrow\frac{2}{A}-\frac{1}{A}=\left(2+2^2+2^3+...+2^{2013}\right)-\left(1+2+2^2+...+2^{2012}\right)\)
\(\Rightarrow\frac{1}{A}=2^{2013}+1\)
\(\Rightarrow A=\frac{1}{2^{2013}+1}\)
D
22 tháng 5 2017
a. \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2013}}\)
\(\Rightarrow3A-A=1-\frac{1}{3^{2014}}\)
\(\Rightarrow2A=1-\frac{1}{3^{2014}}\)
\(\Rightarrow A=\left(1-\frac{1}{3^{2014}}\right):2=\frac{1}{2}-\frac{1}{3^{2014}.2}=\frac{3^{2014}-1}{3^{2014}.2}\)
b.\(B=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2014}}\)
\(\Rightarrow2B=1+\frac{1}{2^2}+....+\frac{1}{2^{2013}}\)
\(\Rightarrow2B-B=1-\frac{1}{2^{2014}}\)
\(\Rightarrow B=1-\frac{1}{2^{2014}}\)
10 tháng 5 2018
\(A=1+\frac{1}{2}+\frac{1}{^{2^2}}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{^{2^2}}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
WH
10 tháng 5 2018
A=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(\Leftrightarrow A=1+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
Đặt \(I=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(2I=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(2I=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(2I-I=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{2}{2^{2011}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(I=1-\frac{1}{1^{2012}}\)
\(\Rightarrow A=1+\left(1-\frac{1}{2^{2012}}\right)\)
\(\Rightarrow A=2-\frac{1}{2^{2012}}\)
Vậy \(A=2-\frac{1}{2^{2012}}\)
HK
1
14 tháng 6 2015
\(\frac{\left(a^3+a^2\right)+\left(a^2-1\right)}{\left(a^3+a^2\right)+\left(a^2+a\right)+\left(a+1\right)}=\frac{a^2\left(a+1\right)+\left(a+1\right)\left(a+1\right)}{a^2\left(a+1\right)+a\left(a+1\right)+\left(a+1\right)}=\frac{\left(a+1\right)\left(a^2+a-1\right)}{\left(a+1\right)\left(a^2+a+1\right)}=\frac{a^2+a-1}{a^2+a-1}\)
ON
23 tháng 3 2018
a/ (x-2)/4 =5+x/2
=> \(\frac{x-2}{4}\) = \(\frac{x+5}{2}\)
=> 2(x-2) = 4(x+5)
2x-4 =4x+20
4x-2x=-20-4
2x=-24
x=-12
\(A=2^2+2^3+2^4+...+2^{2021}\)
\(\Rightarrow2A=2^3+2^4+2^5+....+2^{2022}\)
\(\Rightarrow A=2^{2022}-2^2\)