Tính giá trị biểu thức sau
A= \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + ....... + \(\dfrac{1}{9900}\)
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A = \(\dfrac{1}{12}\)+ \(\dfrac{1}{20}\)+ \(\dfrac{1}{30}\)+...+\(\dfrac{1}{9900}\)
A = \(\dfrac{1}{3\times4}\)+ \(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{99\times100}\)
A = \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A = \(\dfrac{1}{3}\) - \(\dfrac{1}{100}\)
A = \(\dfrac{97}{300}\)
Lời giải:
Gọi tổng trên là $A$
$A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{99.100}$
$=\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}$
$=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}$
$=\frac{1}{3}-\frac{1}{100}=\frac{97}{300}$
Mới thế đã hai năm trôi qua,câu trả lời từ mọi người vẫn KO XUẤT HIỆN.
Ko biết sau này câu trả lời có xuất hiện hay ko...
\(E=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{2}+\dfrac{1}{12}\)
\(E=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{24}\right)\)
\(E=\dfrac{2}{2}+\dfrac{3}{6}+\left(\dfrac{1}{8}+\dfrac{3}{24}\right)\)
\(E=1+\dfrac{1}{2}+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)\)
\(E=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\dfrac{2}{8}\)
\(E=\dfrac{3}{2}+\dfrac{1}{4}\)
\(E=\dfrac{6}{4}+\dfrac{1}{4}\)
\(E=\dfrac{7}{4}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\cdot\dfrac{0}{6}\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\cdot0\)
\(Q=0\)
\(=\dfrac{3}{4}-\dfrac{5}{6}\times\dfrac{7}{24}\times\dfrac{12}{7}=\dfrac{3}{4}-\dfrac{5}{12}=\dfrac{1}{3}\)
\(\dfrac{3}{4}-\dfrac{5}{6}\left(\dfrac{1}{6}+\dfrac{1}{8}\right):\dfrac{7}{12}\)
\(=\dfrac{3}{4}-\dfrac{5}{6}\cdot\dfrac{7}{24}\cdot\dfrac{12}{7}\)
\(=\dfrac{3}{4}-\dfrac{5}{12}\)
\(=\dfrac{4}{12}=\dfrac{1}{3}\)
lần đầu tiên trong đời thấy dấu . là dấu nhân chỉ thấy dấu sao với cả x thôi
\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}.\dfrac{12}{7}\)
\(=\dfrac{2}{3}+\dfrac{7.3.2.2}{3.7.3.2.3}\)
\(=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
TICK CHO MÌNH NHÉ
Giải:
\(\dfrac{2}{3}\) + \(\dfrac{1}{3}\) . (\(-\dfrac{4}{9}\) + \(\dfrac{5}{6}\) ) : \(\dfrac{7}{12}\)
= \(\dfrac{2}{3}\) + \(^{\dfrac{1}{3}}\) . \(\dfrac{7}{18}\) : \(\dfrac{7}{12}\)
= \(\dfrac{2}{3}\) + \(\dfrac{7}{54}\) : \(\dfrac{7}{12}\)
= \(\dfrac{2}{3}\) + \(\dfrac{2}{9}\)
= \(\dfrac{8}{9}\)
\(=\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{7}{18}\right):\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{2}{9}\)
\(=\dfrac{8}{9}\)
\(A=\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{99x100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....\dfrac{1}{99}-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}=\dfrac{99}{100}\)