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A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56
A = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8
A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8
A = 1 + ( -1/2 + 1/2 ) + ( -1/3 + 1/3 ) + ( -1/4 + 1/4 ) + ( -1/5 + 1/5 ) + ( -1/6 + 1/6 ) + ( -1/7 + 1/7 ) - 1/8
A = 1 + 0 + 0 + 0 + 0 + 0 + 0 - 1/8
A = 1 - 1/8
A = 7/8
* Sửa đề tí nhé
B = 3/2 - 5/6 + 7/12 - 9/20 + 11/30 - 13/42 + 15/56
B = 3/1.2 - 5/2.3 + 7/3.4 - 9/4.5 + 11/5.6 - 13/6.7 + 15/7.8
B = 3 - 3/2 - 5/2 - ( -5/3 ) + 7/3 - 7/4 - 9/4 - ( -9/5 ) + 11/5 - 11/6 - 13/6 - ( -13/7 ) + 15/7 - 15/8
B = 3 - 3/2 - 5/2 + 5/3 + 7/3 - 7/4 - 9/4 + 9/5 + 11/5 - 11/6 - 13/6 + 13/7 + 15/7 - 15/8
B = 3 + ( -3/2 - 5/2 ) + ( 5/3 + 7/3 ) + ( -7/4 - 9/4 ) + ( 9/5 + 11/5 ) + ( -11/6 - 13/6 ) + ( 13/7 + 15/7 ) - 15/8
B = 3 + -4 + 4 + -4 + 4 + -4 + 4 - 15/8
B = 3 + 0 + 0 + 0 - 15/8
B = 3 - 15/8
B = 9/8
Lời giải:
$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{x(x+1)}=\frac{3}{8}$
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{x(x+1)}=\frac{3}{8}$
$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{6}-\frac{1}{7}+\frac{1}{x(x+1)}=\frac{3}{8}$
$1-\frac{1}{7}+\frac{1}{x(x+1)}=\frac{3}{8}$
$\frac{6}{7}+\frac{1}{x(x+1)}=\frac{3}{8}$
$\frac{1}{x(x+1)}=\frac{3}{8}-\frac{6}{7}=\frac{-27}{56}$
Kết quả này không phù hợp lắm.
Bạn xem lại đề nhé.
a) \(\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{10}{16}+\dfrac{10}{24}\)
\(=\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{5}{8}+\dfrac{5}{12}\)
\(=\left(\dfrac{3}{8}+\dfrac{5}{8}\right)+\left(\dfrac{7}{12}+\dfrac{5}{12}\right)\)
\(=1+1\)
\(=2\)
b) \(\dfrac{4}{6}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{14}{6}\)
\(=\dfrac{2}{3}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{7}{3}\)
\(=\left(\dfrac{2}{3}+\dfrac{7}{3}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)+\left(\dfrac{17}{9}+\dfrac{1}{9}\right)\)
\(=3+2+2\)
\(=7\)
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(=1-\dfrac{1}{7}\)
\(=\dfrac{6}{7}\)
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\)- \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\) - \(\dfrac{1}{10.11}\) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\) =\(x\)-\(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{11}\) = \(x\) - \(\dfrac{5}{13}\)
\(x-\dfrac{5}{13}=\dfrac{1}{11}\)
\(x\) = \(\dfrac{1}{11}\) + \(\dfrac{5}{13}\)
\(x\) = \(\dfrac{68}{143}\)
tìm x mà lại có y là sao
Y - 1/2 - 1/6 - 1/12 - 1/20 - 1/30 - 1/42 = 1
Đặt A= - 1/2 - 1/6 - 1/12 - 1/20 - 1/30 - 1/42
\(A=-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\right)\)
\(A=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{6.7}\right)\)
\(A=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=-\left(1-\frac{1}{7}\right)\)
\(A=-\frac{6}{7}\).Thay A vào ta có \(Y-\frac{6}{7}=1\Leftrightarrow y=\frac{13}{7}\)