Question

 

Tính nhanh :

 a.  5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90

b.  1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90+109/110

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Answer 1

a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\)

Từ 2 đến 9 có : ( 9 - 2 ) / 1 + 1 = 8 ( số hạng ) => có 8 số 1

\(\Rightarrow8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=8-\frac{2}{5}=\frac{38}{5}\)

b) \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{110}\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)

\(=\left(1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\right)\)

Từ 1 đến 10 có : ( 10 - 1 ) / 1 + 1 = 10 ( số hạng ) => có 10 số 1

\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(=10-\left(1-\frac{1}{11}\right)\)

\(=10-\frac{10}{11}=\frac{100}{11}\)

Answer 2

100/11 nha

Answer 3

câu b bằng 891/110

câu a bằng 76/10

 

Answer 4

a) 56+1112+1920+...+899065​+1211​+2019​+...+9089​

=1−16+1−112+...+1−190=1−61​+1−121​+...+1−901​

=(1+1+1+...+1)−(16+112+...+190)=(1+1+1+...+1)−(61​+121​+...+901​)

=(1+1+1+...+1)−(12⋅3+13⋅4+...+19⋅10)=(1+1+1+...+1)−(2⋅31​+3⋅41​+...+9⋅101​)

Từ 2 đến 9 có : ( 9 - 2 ) / 1 + 1 = 8 ( số hạng ) => có 8 số 1

⇒8−(12−13+13−14+...+19−110)⇒8−(21​−31​+31​−41​+...+91​−101​)

=8−(12−110)=8−(21​−101​)

=8−25=385=8−52​=538​

b) 12+56+1112+...+10911021​+65​+1211​+...+110109​

=1−12+1−16+...+1−1110=1−21​+1−61​+...+1−1101​

=(1+1+1+...+1)−(12+16+...+1110)=(1+1+1+...+1)−(21​+61​+...+1101​)

=(1+1+...+1)−(11⋅2+12⋅3+...+110⋅11)=(1+1+...+1)−(1⋅21​+2⋅31​+...+10⋅111​)

Từ 1 đến 10 có : ( 10 - 1 ) / 1 + 1 = 10 ( số hạng ) => có 10 số 1

⇒10−(1−12+12−13+...+110−111)⇒10−(1−21​+21​−31​+...+101​−111​)

=10−(1−111)=10−(1−111​)

=10−1011=10011=10−1110​=11100​