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\(\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{71}{72}+\dfrac{89}{90}=1-\dfrac{1}{2}+1-\dfrac{1}{6}+1-\dfrac{1}{12}+....+1-\dfrac{1}{90}=1+1+...+1-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)=9-\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+...+\dfrac{1}{9x10}\right)=9-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)=9-\left(1-\dfrac{1}{10}\right)=9-\dfrac{9}{10}=\dfrac{81}{10}\)
ta thay day so tren la day so tu nhien lien tiep bat dau tu 1 va ket thuc la 100 nen co 100 so hang
tong la (100+1) x 100 :2=5050
1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + ... + 301 + 302 (có 302 số; 302 chia 4 dư 2)
= 1 + (2 - 3 - 4 + 5) + (6 - 7 - 8 + 9) + ... + (288 - 299 - 300 + 301) + 302
= 1 + 0 + 0 + ... + 0 + 302
= 1 + 302
= 303
1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + ....... +301 + 302 ( có 302 số : 302 chia 4 dư 2 )
= 1 + ( 2 - 3 - 4 + 5 ) + ( 6 - 7 - 8 + 9 ) + ..... + ( 288 - 299 -300 + 301 ) + 302
= 1 + 0 + 0 + ......... + 0 + 302
= 1 + 302
= 303
(1+19)+(2+18)+(3+17)+(4+16)+(5+15)+(6+14)+(7+13)+(8+12)+(9+11)+20
=20+20+20+20+20+20+20+20+20+20
=20x10
=200
Lời giải:
$\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}$
$=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}$
$=(1+1+....+1)-(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90})$
$=8-(\frac{1}{2\times 3}+\frac{1}{3\times 4}+\frac{1}{4\times 5}+...+\frac{1}{9\times 10})$
$=8-(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10})$
$=8-(\frac{1}{2}-\frac{1}{10})=8-\frac{2}{5}=\frac{38}{5}$