\(\frac{4}{3\times7}+\frac{5}{7\times12}+\frac{1}{12\times13}+\frac{7}{13\times20}+\frac{3}{20\times23}\)
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\(a)\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{-11.13^7}{11^5.13^8}=\frac{-1}{11^4.13}\) (Bạn xem thử xem có sai đề không nhé)
\(c)\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3+1\right)}{2^9.3^{10}}=\frac{2.4}{3}=\frac{8}{3}\)
\(d)\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(5.4+9\right)}=\frac{8}{20+9}=\frac{8}{29}\)
\(a)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{\left(-11\right)\cdot13^7}{11^5\cdot13^8}=\frac{-1}{11^4\cdot13}=\frac{-1}{14641\cdot13}=\frac{-1}{190333}\)
\(c)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\left(3^{10}-3^9\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\cdot2}{2^9\cdot3^{10}}=\frac{2\cdot2}{3}=\frac{4}{3}\)
Bài 1 :
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(S=\frac{1}{1}-\frac{1}{2011}=\frac{2010}{2011}\)
Bài 2 :
\(S=\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+...+\frac{1}{58}-\frac{1}{61}\)
\(S=\frac{1}{10}-\frac{1}{61}=\frac{51}{610}\)
Bài 3 :
\(3S=\frac{3}{4\times7}+\frac{3}{7\times11}+...+\frac{3}{19\times22}\)
\(3S=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{19}-\frac{1}{22}\)
\(3S=\frac{1}{4}-\frac{1}{22}\)
\(S=\frac{18}{88}\div3=\frac{6}{88}\)
1 x 3 x 5 + 2 x 6 x 10 + 4 x 10 x 12 + 7 x 21 x 35 / 1 x 5 x 7 + 2 x 10 x 14 + 4 x 20 x 28 + 7 x 35 x 49
= 3 / 7 + 6 / 14 + 10 x 2 x 6 / 10 x 2 x 28 + 21 / 49
= 3 / 7 + 6 / 14 + 6 / 28 + 21 / 49
= 3 / 7 + 6 / 14 + 3 / 14 + 3 / 7
= ( 3 / 7 + 3 / 7 ) + ( 6 / 14 + 3 / 14 )
= 6 / 7 + 9 14
= 12 / 14 + 9 / 14
= 21 / 14
nho hem
lm dau tien lun do
dung 100% nha
thế này :
= \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{11.13}\right)\)
= \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{11}-\frac{1}{13}\right)\)
= \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)
= \(\frac{1}{2}.\frac{10}{39}\)
= \(\frac{5}{39}\)
Vậy kq = \(\frac{5}{39}\)
\(\frac{20}{69}\)
\(\frac{4}{3x7}\)+ \(\frac{5}{7x12}\)+ \(\frac{1}{12x13}\)+ \(\frac{7}{13x20}\)+ \(\frac{3}{20x23}\)
= \(\frac{4}{3}+5\)+\(1+7+\frac{3}{13}\)
=\(4+5+1+7+\frac{1}{13}\)
=\(17+\frac{1}{13}\)
=\(\frac{17}{1}+\frac{1}{13}=\frac{221+1}{13}=\frac{222}{13}\)