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\(\dfrac{8}{9}:\dfrac{3}{7}=\dfrac{56}{27}\\ \dfrac{8}{9}+\dfrac{2}{5}=\dfrac{58}{45}\\ \dfrac{7}{8}-\dfrac{1}{3}=\dfrac{13}{24}\\ \dfrac{3}{10}\times\dfrac{1}{6}=\dfrac{1}{20}\\ 1\dfrac{2}{7}+6\dfrac{5}{6}=\dfrac{9}{7}+\dfrac{41}{6}=\dfrac{341}{42}\\ 5\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{23}{4}-\dfrac{1}{5}=\dfrac{111}{20}\\ 6\dfrac{2}{9}:4\dfrac{7}{10}=\dfrac{56}{9}:\dfrac{47}{10}=\dfrac{560}{423}\\ \dfrac{5}{3}+\dfrac{3}{2}-\dfrac{7}{6}=2\)
= 1 x 3 + 2 x 3 + 3 x 3 + 4 x 3 + ...+ 9 x 3
= 3 x ( 1 + 2 + 3 + 4 + ...+ 9)
= 3 x 45
= 135
1+2+3+4+5+6+7+8+9+1+2+3+4+5+6+7+8+9+1+2+3+4+5+6+7+8+9+1028=(1+2+3+4+5+6+7+8+9)x3+1028=45x3+1028=135+1028=1163
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
a)\(2-3+5-7+9-11+13-15+17=\left(2+5+9+13+17\right)-\left(3+7+11+15\right)\)
\(=46-36=10\)
b)\(\frac{1}{1.2}+\frac{1}{2.3}+...............+\frac{1}{8.9}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.................+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{1}-\frac{1}{9}=\frac{9}{9}-\frac{1}{9}=\frac{8}{9}\)
Áp dụng \(\frac{1}{n.\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
Chúc bạn học tốt
Đề là tính bằng cách hợp lý đúng ko bạn
a, 2-3+5-7+9-11+13-15+17
= (5+13) - (3+15) + (2+9-11) + (17-7)
= 18 - 18 + 0 +10
= 10
b, \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)