Решите номера 93 и 135 (а), пожалуйста.

Question 1

Question 2

93
m/n=1,2 m=1,2n
а)m/n=10/12=5/6
б)-5m/-3n=-5/3*1,2=-5*0,4=-2
в)3+2n/m=3+2*5/6=3+1 2/3=4 2/3
г)(3m-2n)/(2m+n)=(3,6n-2n)/(2,4n+n)=1,6n/3,4n=16/34=8/17
135(a)
1/5*6+1/6*7+....+1/19*20=(1/5-1/6)+(1/6-1/)+...+(1/19-1/20)=
=1/5-1/6+1/6-1/7+1/7-....+1/19-1/20=1/5-1/20=(4-1)/20=3/20=0,3

Question 3

Решите задачу:

$\frac{m}{n} =1,2\quad \\\\ \frac{n}{m} =1: \frac{m}{n} = \frac{1}{1,2} = \frac{10}{12} = \frac{5}{6} \\\\- \frac{5m}{3n} =- \frac{5}{3} \cdot \frac{m}{n} =- \frac{3}{5} \cdot 1,2=- \frac{3\cdot 1,2}{5} =- \frac{3,6}{5} =- \frac{36}{50}=- \frac{18}{25} = -0,72\\\\3+ \frac{2n}{m}=3+2\cdot \frac{n}{m} =3+2\cdot \frac{5}{6} =3+ \frac{5}{3} = \frac{9+5}{3} = \frac{14}{3} =4 \frac{2}{3}$

$\frac{3m-2n}{2m+n} = \frac{n(3\cdot \frac{m}{n} -2)}{n(2\cdot \frac{m}{n} +1)} = \frac{3\cdot 1,2-2}{2\cdot 1,2+1} =\frac{1,6}{3,4} = \frac{16}{34} = \frac{8}{17}$

$2a)\; \; \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$

$\frac{1}{5\cdot 6} + \frac{1}{6\cdot 7} + \frac{1}{7\cdot 8} +...+ \frac{1}{18\cdot 19} + \frac{1}{19\cdot 20} =\\\\= (\frac{1}{5} -\frac{1}{6} )+ (\frac{1}{6} - \frac{1}{7} )+( \frac{1}{7} - \frac{1}{8})+...+( \frac{1}{18} - \frac{1}{19} ) +( \frac{1}{19} - \frac{1}{20} )=\\\\= \frac{1}{5} - \frac{1}{20} = \frac{4-1}{20} = \frac{3}{20}$

$b)\; \; \frac{1}{(2n-1)(2n+1)} = \frac{1}{2}(\frac{1}{2n-1} -\frac{1}{2n+1} )$

$\frac{1}{9\cdot 11} + \frac{1}{11\cdot 13} + \frac{1}{13\cdot 15} +...+ \frac{1}{21\cdot 23} + \frac{1}{23\cdot 25} =\\\\= \frac{1}{2} (\frac{1}{9} - \frac{1}{11} )+ \frac{1}{2} ( \frac{1}{11} - \frac{1}{13} )+\frac{1}{2}( \frac{1}{13} - \frac{1}{15} )+...+ \frac{1}{2} ( \frac{1}{21} - \frac{1}{23} )+\\\\+\frac{1}{2}( \frac{1}{23} - \frac{1}{25} )=\frac{1}{2}(\frac{1}{9}-\frac{1}{25})=\frac{1}{2}\cdot \frac{25-9}{9\cdot 25} = \frac{8}{225}$