Đáp án:
$9.5)
B=\frac{10}{39}\\
9.6)
C=\frac{18}{19}$
Giải thích các bước giải:
$9.5)
B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\\
=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\\
=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\\
=\frac{1}{3}-\frac{1}{13}\\
=\frac{13}{39}-\frac{3}{39}\\
=\frac{10}{39}\\
9.6)
C=\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\\
=\frac{1}{2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}\\
=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+\frac{1}{13}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}\\
=1-\frac{1}{19}\\
=\frac{18}{19}\\
9.7)
D=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\\
<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\\
=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\\
=1-\frac{1}{10}\\
=\frac{9}{10}\\
<\frac{10}{10}=1$
