Rút gọn A
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{50}}\)
K
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H9
HT.Phong (9A5)
CTVHS
18 tháng 8 2023
a) \(a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}\cdot a^{\dfrac{7}{6}}=a^{\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{7}{6}}=a^2\)
b) \(a^{\dfrac{2}{3}}\cdot a^{\dfrac{1}{4}}:a^{\dfrac{1}{6}}=a^{\dfrac{2}{3}+\dfrac{1}{4}-\dfrac{1}{6}}=a^{\dfrac{3}{4}}\)
c) \(\left(\dfrac{3}{2}a^{-\dfrac{3}{2}}\cdot b^{-\dfrac{1}{2}}\right)\left(-\dfrac{1}{3}a^{\dfrac{1}{2}}b^{\dfrac{2}{3}}\right)=\left(\dfrac{3}{2}\cdot-\dfrac{1}{3}\right)\left(a^{-\dfrac{3}{2}}\cdot a^{\dfrac{1}{2}}\right)\left(b^{-\dfrac{1}{2}}\cdot b^{\dfrac{2}{3}}\right)\)
\(=-\dfrac{1}{2}a^{-1}b^{-\dfrac{1}{3}}\)
KK
2
YN
4 tháng 2 2020
\(A=\frac{\frac{3}{2}+\frac{2}{5}+\frac{1}{10}}{\frac{3}{2}+\frac{2}{3}+\frac{1}{12}}\)
\(\Rightarrow A=\frac{\frac{15}{10}+\frac{4}{10}+\frac{1}{10}}{\frac{18}{12}+\frac{8}{12}+\frac{1}{12}}=\frac{\frac{20}{10}}{\frac{27}{12}}=\frac{2}{\frac{9}{4}}=2:\frac{9}{4}=2.\frac{4}{9}=\frac{8}{9}\)
! Ko bt có đúng ko nx @@@
~ Học tốt
# Chiyuki Fujito
WH
8 tháng 4 2018
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(\Rightarrow A=1+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(2B=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\right)\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(B=1-\frac{1}{2^{2012}}\)
\(\Rightarrow A=1+\left(1-\frac{1}{2^{2012}}\right)\)
\(\Rightarrow A=2-\frac{1}{2^{2012}}\)
MT
3
9 tháng 7 2020
Trả lời
\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.\left(\frac{1}{a^2+2a+1}-\frac{1}{a^2-1}\right)\) \(\left(a\ge0.a\ne1\right)\)
\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.\left[\frac{1}{\left(a+1\right)^2}-\frac{1}{\left(a-1\right).\left(a+1\right)}\right]\)
\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.\left[\frac{a-1-a-1}{\left(a+1\right)^2.\left(a-1\right)}\right]\)
\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}.0\)
\(B=\frac{1}{a+1}\)
Vậy \(B=\frac{1}{a+1}\)
9 tháng 7 2020
\(B=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\left(\frac{1}{a^2+2a+1}-\frac{1}{a^2-1}\right)ĐK\left(a\ge0;a\ne1\right)\)
\(=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\left(\frac{a^2-1}{\left(a^2+1\right)\left(a^2-1\right)}-\frac{a^2+1}{\left(a^2-1\right)\left(a^2+1\right)}\right)\)
\(=\frac{1}{a+1}+\frac{a-a^3}{a^2+1}\left(\frac{a^2-1-a^2-1}{\left(a^2+1\right)\left(a^2-1\right)}\right)\)
\(=\frac{1}{a+1}\)
Vậy \(B=\frac{1}{a+1}\)
T
2
11 tháng 5 2019
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2012}}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2011}}\)
\(A=2-\frac{1}{2^{2012}}\)
14 tháng 1 2017
\(A=\frac{\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right)}{\left(\frac{3}{2}-\frac{2}{3}+\frac{1}{12}\right)}\)
\(A=\frac{\left(\frac{15}{10}-\frac{4}{10}+\frac{1}{10}\right)}{\left(\frac{18}{12}-\frac{8}{12}+\frac{1}{12}\right)}\)
\(A=\frac{\frac{6}{5}}{\frac{11}{12}}=\frac{6}{5}:\frac{11}{12}=\frac{6}{5}\times\frac{12}{11}\)
\(A=\frac{72}{55}\)
13 tháng 4 2017
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
Nên \(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
Suy ra \(2A-A=2-\frac{1}{2^{2012}}\)hay \(A=2-\frac{1}{2^{2012}}\)
Vậy \(A=2-\frac{1}{2^{2012}}\)
13 tháng 4 2017
\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\)
=>\(A-\frac{1}{2}A=\left(1+\frac{1}{2}+..+\frac{1}{2^{2012}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\right)\)
=>\(\frac{1}{2}A=1-\frac{1}{2^{2013}}\)
=>\(A=2-\frac{1}{2^{2012}}\)
Cô mình chữa bài này rồi nên bạn cứ yên tâm
Ta có : \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{50}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{49}}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{50}}\)
\(\Rightarrow A=1-\frac{1}{2^{50}}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}\)
\(A=\frac{2^{50}-1}{2^{50}}.\)