Tính S = 1 / 5 x 9 + 1 / 9 x 13 + 1 / 13 x 17 + ....... + 1 / 41 x 45
Giải thích giúp tôi nếu trong phép tính có 1 / 4
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\(4.S=\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\)
\(4S=\frac{1}{5}-\frac{1}{6}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)
\(4S=\frac{1}{5}-\frac{1}{45}=\frac{8}{45}\)
\(S=\frac{8}{45}:4=\frac{2}{45}\)
\(S=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\)
\(\Rightarrow4S=\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\)
\(\Rightarrow4S=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\)
\(\Rightarrow4S=\frac{1}{5}-\frac{1}{45}=\frac{8}{45}\Rightarrow S=\frac{8}{45}:4=\frac{2}{45}\)
~ Hok tốt ~
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
Bài 2 :
a, \(x=\dfrac{3}{5}-\dfrac{7}{8}=\dfrac{24-30}{40}=-\dfrac{6}{40}=-\dfrac{3}{20}\)
b, \(2x-1=-2\Leftrightarrow x=-\dfrac{1}{2}\)
1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
\(S=\dfrac{1}{5\cdot9}+\dfrac{1}{9\cdot13}+\dfrac{1}{13\cdot17}+....+\dfrac{1}{41\cdot45}\)
\(S=\dfrac{1}{4}\cdot\left(\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+\dfrac{4}{13\cdot17}+....+\dfrac{4}{41\cdot45}\right)\)
\(S=\dfrac{1}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}+....+\dfrac{1}{41}-\dfrac{1}{45}\right)\)
\(S=\dfrac{1}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{45}\right)\)
\(S=\dfrac{1}{4}\cdot\dfrac{8}{45}=\dfrac{2}{45}\)
Anh ơi :) giải thích cho em là lấy 1/4 ở đâu với ? Em học ngu