TÍNH NHANH TỔNG SAU :
1/42 +1/30 +1/20 +1/12 +1/6
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Đặt \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+......+\frac{1}{56}\)
\(\Rightarrow A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{7.8}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow A=1-\frac{1}{8}\)
\(\Rightarrow A=\frac{7}{8}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+...+\left(1-\dfrac{1}{90}\right)\\ =\left(1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\\ =9-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\\ =9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =9-\left(1-\dfrac{1}{10}\right)=9-\dfrac{9}{10}=\dfrac{81}{10}\)
a/ \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}=\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}=\)
= 1-1/8 = 7/8
b/ = (17/9+1/9)+(19/13+7/13)+(14/6+10/6) = 18/9 + 26/13 + 24/6 = 2+2+4 = 8
1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90
=(1-1/2)+(1-1/6)+(1-1/12)+(1-1/30)+(1-1/42)+(1-1/56)+(1-1/72)+(1-1/90)
=(1+1+1+1+1+1+1+1+1)-(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)
Ta có : A=1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90
A=1/1.2+1/2.3+1/3.4=1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
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+1/8-1/9+1/9-1/10
A=1-1/10
Thay vào ta có
=9-9/10
=81/10
Đầy đủ luôn nhé
\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{38}{5}\)
Theo đề ra, ta có:
\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}-\dfrac{1}{6}\right)+\left(\dfrac{1}{6}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{8}\right)\)
\(=\dfrac{1}{2}+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+\left(-\dfrac{1}{6}+\dfrac{1}{6}\right)+\left(-\dfrac{1}{7}+\dfrac{1}{7}\right)-\dfrac{1}{8}\)
\(=\dfrac{1}{2}+0+0+0+0+0-\dfrac{1}{8}\)
\(=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{4}{8}-\dfrac{1}{8}=\dfrac{3}{8}\)
Chúc bạn học tốt!
Đặt \(A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(A=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}\)
\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
\(A=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)
1/42 + 1/30 +1/20 + 1/12 + 1/6
= 1/6+ 1/12 + 1/20 + 1/30 + 1/42
= 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7
= 1/2 - 1/3 + 1/3 -1/4 + 1 /4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7
= 1/2 - 1/7
= 5/14
\(\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}=\frac{1}{6\times7}+\frac{1}{5\times6}+\frac{1}{4\times5}+\frac{1}{3\times4}+\frac{1}{2\times3}\)
\(=\frac{1}{6}-\frac{1}{7}+\frac{1}{5}-\frac{1}{6}+\frac{1}{4}-\frac{1}{5}+\frac{1}{3}-\frac{1}{4}+\frac{1}{2}-\frac{1}{3}\)
\(=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
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A=1/6 +1/12 +1/20 +1/30 +1/42
=1/2.3 +1/3.4 +1/4.5 +1/5.6 +1/6.7
=1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7
=1/2 - 1/7 = 5/14