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TQ
0
10 tháng 1 2022
\(=\dfrac{2^{19}\cdot3^9-3\cdot3^8\cdot2^{18}\cdot5}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}=\dfrac{-3^{10}\cdot2^{18}}{2^{19}\cdot3^9\cdot7}=-\dfrac{3}{14}\)
18 tháng 9 2021
Bài 1:
\(A=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+\dfrac{9}{16.25}+\dfrac{11}{25.36}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{36}\)
\(=1-\dfrac{1}{36}=\dfrac{35}{36}\)
\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=1-\dfrac{1}{103}=\dfrac{102}{103}\)
\(C=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}+\dfrac{15}{31.46}+\dfrac{18}{46.64}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{64}\)
\(=1-\dfrac{1}{64}=\dfrac{63}{64}\)
Bài 2:
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\left(đpcm\right)\)
TQ
1
GD
0
KS
0
LG
0
NT
1
\(\dfrac{3^{12}+3^{15}}{1+3^3}\)
\(=\dfrac{3^{12}\left(1+3^3\right)}{1+3^3}\)
\(=3^{12}\)
\(\dfrac{3^{12}+3^{15}}{1+3^3}=\dfrac{3^{12}\left(1+3^3\right)}{1+3^3}=3^{12}\)