các bạn ơi giúp mình câu này nhé 2x5:4x9
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\(\dfrac{1}{1.3}+\dfrac{1}{2.3}+\dfrac{1}{2.5}+\dfrac{1}{3.5}+\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.9}\)
\(=\dfrac{1}{1.3}+\dfrac{1}{3.2}+\dfrac{1}{2.5}+\dfrac{1}{5.3}+\dfrac{1}{3.7}+\dfrac{1}{7.4}+\dfrac{1}{4.9}\)
\(=\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right):\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right):\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{9}\right):\dfrac{1}{2}\)
\(=\dfrac{7}{18}:\dfrac{1}{2}\)
\(=\dfrac{7}{9}\)
\(=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{1999\cdot2004}\right)\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{1999}-\dfrac{1}{2004}\right)\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{2004}\right)=\dfrac{1}{5}\cdot\dfrac{125}{501}=\dfrac{25}{501}\)
\(\dfrac{2\times11\times3}{4}\) x \(\dfrac{9}{121}\) = \(\dfrac{2.11.3.9}{2.2.11.11}\) = \(\dfrac{2.11}{2.11}\). \(\dfrac{3.9}{2.11}\)= \(\dfrac{27}{22}\)
Đặt \(A=\frac{3}{1\cdot4}+\frac{5}{4\cdot9}+\frac{7}{9\cdot16}+\frac{9}{16\cdot25}+\frac{11}{25\cdot36}\)
\(A=\frac{4-1}{1\cdot4}+\frac{9-4}{4\cdot9}+\frac{16-9}{9\cdot16}+\frac{25-16}{16\cdot25}+\frac{36-25}{25\cdot36}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{25}+\frac{1}{25}-\frac{1}{36}\)
\(\Rightarrow A=1-\frac{1}{36}=\frac{35}{36}\)
= 1-\(\frac{1}{4}\)+ \(\frac{1}{4}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{16}\)+\(\frac{1}{16}\)-\(\frac{1}{25}\)+\(\frac{1}{25}\)-\(\frac{1}{36}\)=1-\(\frac{1}{36}\)=35/36
120+[1+2+3...+9+100]x[4x9-36]
=120+1+2+3...+9+100]x0
=120+0
=120
\(S=\frac{101}{120}+\frac{1}{2.3}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(S=\frac{101}{120}+\frac{1}{6}\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{19-18}{18.19}+\frac{20-19}{19.20}\right)\)
\(S=\frac{101}{120}+\frac{1}{6}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(S=\frac{101}{120}+\frac{1}{6}\left(1-\frac{1}{20}\right)=\frac{101}{120}+\frac{19}{120}=\frac{120}{120}=1\)
2 x 5 : 4 x 9
= 10 : 4 x 9
= \(\frac52\) x 9
= \(\frac{45}{2}\)