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17 tháng 2 2018
1) \(+2x+3y⋮17\)
\(\Rightarrow26x+39y⋮17\)
\(\Rightarrow\left(9x+5y\right)+17x+34y⋮17\)
Mà \(17x+34y⋮17\)
\(\Rightarrow9x+5y⋮17\)
\(+9x+5y⋮17\)
\(\Rightarrow36x+20y⋮17\)
\(\Rightarrow\left(2x+3y\right)+34x+17y⋮17\)
Mà \(34x+17y⋮17\)
\(\Rightarrow2x+3y⋮17\)
KT
23 tháng 10 2017
A = \(\frac{1}{2}\)\(-\)\(\frac{1}{2^2}\)\(+\)\(\frac{1}{2^3}\)\(-\)\(\frac{1}{2^4}\)\(+\)........... \(+\)\(\frac{1}{2^{99}}\)\(-\)\(\frac{1}{2^{100}}\)
2A = 1 - \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)- \(\frac{1}{2^3}\)+.........+ \(\frac{1}{2^{98}}\)- \(\frac{1}{2^{99}}\)
2A + A =( 1 - \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)- \(\frac{1}{2^3}\)+.........+ \(\frac{1}{2^{98}}\)- \(\frac{1}{2^{99}}\)) \(+\)( \(\frac{1}{2}\)\(-\)\(\frac{1}{2^2}\)\(+\)\(\frac{1}{2^3}\)\(-\)\(\frac{1}{2^4}\)\(+\)........... \(+\)\(\frac{1}{2^{99}}\)\(-\)\(\frac{1}{2^{100}}\))
3A = 1 \(-\) \(\frac{1}{2^{100}}\)
\(\Rightarrow\)A = \(\frac{1-\frac{1}{2^{100}}}{3}\)= \(\frac{1}{3}\)
18 tháng 12 2018
a) \(\frac{1}{7}+\frac{6}{7}:\frac{3}{7}\)
\(=\frac{1}{7}+\frac{6}{7}.\frac{7}{3}\) (nhân nghịch đảo)
\(=\frac{1}{7}+2\)
\(=\frac{15}{7}\)
b) \(\frac{4}{5}-\frac{1}{5}.\left(-3\right)\)
\(=\frac{4}{5}-\left(-\frac{3}{5}\right)\)
\(=\frac{7}{5}\)
c) \(\frac{3}{7}+\left(\frac{-5}{2}\right)-\left(-\frac{3}{5}\right)\)
\(=\frac{3}{7}-\left(-\frac{5}{2}\right)+\frac{3}{5}\)
\(=\frac{30}{70}+\frac{175}{70}+\frac{42}{70}\)
\(=\frac{30+175+42}{70}\)
\(=\frac{247}{70}\)
d) viết lại đề hộ mình nhé
18 tháng 3 2018
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}\)
\(B=\frac{1}{100}\)
C
21 tháng 8 2015
Ta có: \(y=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}\Leftrightarrow3y=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{98}}\)
\(\Leftrightarrow3y-y=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{99}\right)\)
\(\Leftrightarrow2y=1-\frac{1}{3^{99}}<1\Leftrightarrow y<\frac{1}{2}\)
Phần b tương tự
tick cho mình nha
C
3 tháng 2 2019
Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)
\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)
\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
\(A=\frac{B}{6}=\frac{100}{2}=50\)
Vậy \(A=50\)
30 tháng 3 2017
Ta có 99/1+98/2+97/3+...+1/99=(98/2+1)+(97/3+1)+...+(1/99+1)+1
=100/2+100/3+...+100/99+100/100
=100(1/2+1/3=1/4+1/5+...+1/99+1/100)
Vậy (1/2+1/3+...+1/100)/((99/1+98/2+...+1/99)=1/100
\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(\Rightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{9^{97}}+\frac{1}{3^{98}}\)
\(\Rightarrow3C-C=1-\frac{1}{3^{98}}\)
\(\Rightarrow C=\frac{1-\frac{1}{3^{98}}}{2}\)
Nhầm một chút ==
\(3C-C=2C=1-\frac{1}{3^{99}}\)
\(\Rightarrow C=\frac{1-\frac{1}{3^{99}}}{2}\)