\(1\frac{1}{2}+2\frac{1}{6}+3\frac{1}{12}+4\frac{1}{20}+5\frac{1}{30}+6\frac{1}{42}+7\frac{1}{56}+8\frac{1}{72}+9\frac{1}{90}+\frac{1}{10}\)\(\frac{1}{10}=?\)
Tính nhanh:
A=\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
B=\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}\)
A = \(\frac{-79}{90}\)
B = \(\frac{8}{9}\)
\(\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{3}\)
\(\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
9/1-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2=0/4
Giải :
ta có
\(\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
=\(\frac{9}{10}-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{9\times10}\right)\)
=\(\frac{9}{10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
=\(\frac{9}{10}-\left[1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\right]\)
=\(\frac{9}{10}-\left(1-\frac{1}{10}\right)\)
=\(\frac{9}{10}-1+\frac{1}{10}=0\) (Mong online math ks cho mình nhé)
Tính nhanh:
S=\(\frac{3}{4}-0,25-[\frac{7}{3}+(\frac{-9}{2})]-\frac{5}{6}\)
X=\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(S=\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(\frac{-9}{2}\right)\right]-\frac{5}{6}\)
\(S=\frac{3}{4}-\frac{1}{4}-\left[\frac{14}{6}+\left(\frac{-27}{6}\right)\right]-\frac{5}{6}\)
\(S=\frac{1}{2}-\left(\frac{-13}{6}\right)-\frac{5}{6}\)
\(S=\frac{3}{6}-\left(\frac{-13}{6}\right)-\frac{5}{6}\)
\(S=\frac{11}{6}\)
Thực hiện phép tính:
\(A=\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(A=\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{6}-\frac{1}{2}\)
\(A=\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(A=\frac{9}{10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=\frac{9}{10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\frac{9}{10}-\left(1-\frac{1}{10}\right)\)
\(A=\frac{9}{10}-\frac{9}{10}=0\)
\(A=\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{6}-\frac{1}{2}\)
\(\Leftrightarrow A=\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(\Leftrightarrow A=\frac{9}{10}-\frac{9}{10}\)
\(\Leftrightarrow A=0\)
\(\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
Ta có : \(\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{9}{10}-\left(\frac{1}{90}+\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=\frac{9}{10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{9.10}\right)\)
\(=\frac{9}{10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{9}{10}-\left(1-\frac{1}{10}\right)\)
\(=\frac{9}{10}-\frac{9}{10}=0\)
Tính :
\(\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
=\(\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{9}{10}-(\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1})\)
=\(\frac{9}{10}-(\frac{1}{10}+\frac{1}{9}-\frac{1}{9}+\frac{1}{8}-\frac{1}{8}+\frac{1}{7}-\frac{1}{7}+\frac{1}{6}-\frac{1}{6}+\frac{1}{5}-\frac{1}{5}+\frac{1}{4}-\frac{1}{4}+\frac{1}{3}-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+\frac{1}{1})\)
= \(\frac{9}{10}-\left(\frac{1}{10}+\frac{1}{1}\right)\)
=\(\frac{9}{10}-\left(\frac{1}{10}+\frac{10}{10}\right)\)
=\(\frac{9}{10}-\frac{11}{10}\)
= \(\frac{-2}{10}=-\frac{1}{5}\)
# Chúc bạn học tốt #
\(\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{9}{10}-\left(\frac{1}{9.10}+\frac{1}{8.9}+\frac{1}{7.8}+...+\frac{1}{2.3}+\frac{1}{1.2}\right)\)
\(=\frac{9}{10}-\left(\frac{1}{9}-\frac{1}{10}+\frac{1}{8}-\frac{1}{9}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{2}-\frac{1}{3}+\frac{1}{1}-\frac{1}{2}\right)\)
\(=\frac{9}{10}-\left(1-\frac{1}{10}\right)\)
\(=0\)
Hi Hi!
B=\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+10}\)
C=\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
Ta có công thức : \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)
\(\Rightarrow B=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+10}\)
\(=\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+10\right)10}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{10.11}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{11}\right)=2.\frac{9}{22}=\frac{9}{11}\)
Tìm x biết :
a) \(\frac{1}{2013}\times x+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2012\times2013}=2\)
b)\(2x+\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}+\frac{73}{72}+\frac{91}{90}=10\)