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\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{9}{20}\)
1/1x2 + 1/2×3 + 1/3×4 + 1/4×5 +....+ 1/18×19 + 1/19×20
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +....+ 1/18 - 1/19 + 1/19 - 1/20
= 1 - 1/20
= 19/20
= 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/19-1/20
=1/2-1/20
=10/20-1/20
=9/20
\(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{19\times20}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=\dfrac{1}{2}-\dfrac{1}{20}\)
\(=\dfrac{9}{20}\)
= 1/2-1/3+1/3-1/4+1/4-1/5+.........+ 1/19-1/20
= 1/2-1/20
= 9/20
ok bn
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{9}{20}\)
\(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ \(\frac{1}{4x5}\)+ ... + \(\frac{1}{18x19}\)+ \(\frac{1}{19x20}\)
= \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+ ... + \(\frac{1}{18}\)- \(\frac{1}{19}\)+ \(\frac{1}{19}\)- \(\frac{1}{20}\)
= \(\frac{1}{2}\)- \(\frac{1}{20}\)
= \(\frac{18}{40}\)= \(\frac{9}{20}\)
b) \(29\times87-29\times23+64\times71=29\times\left(87-23\right)+64\times71\)
\(=29\times64+64\times71=64\times\left(29+71\right)=64\times100=6400\)
c) \(\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{19\times20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{9}{20}\)
B = 29 x ( 87 - 23 ) + 64 x 71
B = 29 x 64 + 64 x 71
B = 64 x ( 29 + 71 )
B = 64 x 100
B = 6400
D = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)\(\frac{1}{20}\)
D = \(\frac{1}{2}-\frac{1}{20}\)
D = \(\frac{9}{20}\)
Ta dễ dàng nhận thấy: \(\frac{1}{2\times3}=\frac{3-2}{2\times3}=\frac{3}{2\times3}-\frac{2}{2\times3}=\frac{1}{2}-\frac{1}{3}\).
Vậy, ta có thể tính dãy này như sau:
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{19\times20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
Ta gạch đi những phân số giống nhau và bằng nhau, Ta còn \(\frac{1}{2}\)và \(\frac{1}{20}\). Vậy từ đó ta có:
\(=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{18.19}+\)\(\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}=\frac{10-1}{20}=\frac{9}{20}\)
Ta thấy: \(\frac{1}{1.2}=\frac{2-1}{1.2}=\frac{2}{1.2}-\frac{1}{1.2}=1-\frac{1}{2}\); \(\frac{1}{2.3}=\frac{3-2}{2.3}=\frac{3}{2.3}-\frac{2}{2.3}=\frac{1}{2}-\frac{1}{3}\)
Tương tự với các phân số khác
Cho A=2/1x2 + 2/2x3 + 2/3x4 + 2/4x5 + ... + 2/19x20
=> \(A=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(A=2\left(1-\frac{1}{20}\right)=2.\frac{19}{20}=\frac{19}{10}=1,9\)
Chú ý dấu chấm là dấu nhân
\(\frac{2}{1\times2}+\frac{2}{2\times3}+...+\frac{2}{19\times20}\)
\(=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2\times\left(1-\frac{1}{20}\right)=2\times\frac{19}{20}=\frac{19}{10}\)
có ai giúp mình ko
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