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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 :
\(3S=\frac{3}{4\times7}+\frac{3}{7\times11}+...+\frac{3}{19\times22}\)
\(3S=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{19}-\frac{1}{22}\)
\(3S=\frac{1}{4}-\frac{1}{22}\)
\(S=\frac{18}{88}\div3=\frac{6}{88}\)
\(\left(\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{17\times19}\right)\times114-0,2\left(x-1\right)=10\)
\(\Rightarrow\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\right)\right]\times114-0,2x+0,2=10\)
\(\Rightarrow\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{19}\right)\right]\times114+0,2-0,2x=10\)
\(\Rightarrow\frac{8}{57}\times114+0,2-0,2x=10\Rightarrow16+0,2-0,2x=10\)
\(\Rightarrow16,2-0,2x=10\Rightarrow0,2x=16,2-10\Rightarrow0,2x=6,2\Rightarrow x=31\)
\(a)\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{-11.13^7}{11^5.13^8}=\frac{-1}{11^4.13}\) (Bạn xem thử xem có sai đề không nhé)
\(c)\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3+1\right)}{2^9.3^{10}}=\frac{2.4}{3}=\frac{8}{3}\)
\(d)\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(5.4+9\right)}=\frac{8}{20+9}=\frac{8}{29}\)
\(a)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{\left(-11\right)\cdot13^7}{11^5\cdot13^8}=\frac{-1}{11^4\cdot13}=\frac{-1}{14641\cdot13}=\frac{-1}{190333}\)
\(c)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\left(3^{10}-3^9\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\cdot2}{2^9\cdot3^{10}}=\frac{2\cdot2}{3}=\frac{4}{3}\)
Đặt \(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{17x19}\)
=>\(2xA=2x\left(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{17x19}\right)\)
=>\(2xA=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{17x19}\)
=>\(2xA=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)
=>\(2xA=1-\frac{1}{19}=\frac{18}{19}\)
=>\(A=\frac{18}{19}:2=\frac{9}{19}\)
(\(\frac{1}{1}-\frac{1}{3}\left(\right)+\left(\right)\frac{1}{3}-\frac{1}{5}\left(\right)+\left(\right)\frac{1}{5}-\frac{1}{7}\left(\right)+....+\left(\right)\frac{1}{17}-\frac{1}{19}\left(\right)\)\(\frac{1}{19}\)
\(\frac{1}{1}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+....+\left(\frac{1}{17}-\frac{1}{17}\right)-\frac{1}{19}\)
\(\frac{1}{1}-\frac{1}{19}=\frac{18}{19}\)
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 = \(\frac{5}{39}\)
=\(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+....+\frac{2}{11}-\frac{2}{13}\)
=2-\(\frac{2}{13}\)=\(\frac{24}{13}\)
2/
a) \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)
\(=\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{17}-\frac{1}{21}\right)\)
\(=1-\frac{1}{21}=\frac{20}{21}\)
b) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot..\cdot\frac{2016}{2017}\)
\(=\frac{1}{2017}\)
c) \(A=2000-5-5-5-..-5\)(có 200 số 5)
\(A=2000-\left(5\cdot200\right)\)
\(A=2000-1000\)
\(A=1000\)
