Đáp án:
a) Trên cùng 1 mp bờ OM có góc MOP < góc MON
=> tia OP nằm giữa 2 tia OM và ON
$\begin{array}{l}
\Rightarrow \widehat {MOP} + \widehat {NOP} = \widehat {MON}\\
\Rightarrow \widehat {NOP} = {130^0} - {45^0} = {85^0}
\end{array}$
b) Góc NOP và góc NOA là 2 góc kề bù
=> tia OA là tia đối của tia OP
$\begin{array}{l}
B = \left( {7\dfrac{3}{5} + 3\dfrac{5}{{11}}} \right) - 5\dfrac{3}{5}\\
= 7 + \dfrac{3}{5} + 3 + \dfrac{5}{{11}} - 5 - \dfrac{3}{5}\\
= 5 + \dfrac{5}{{11}}\\
= \dfrac{{60}}{{11}}\\
C = \dfrac{{ - 15}}{{17}}.\dfrac{5}{{16}} + \dfrac{{11}}{{16}}.\dfrac{{ - 15}}{{17}} + 2\dfrac{{15}}{{17}}\\
= \dfrac{{ - 15}}{{17}}.\left( {\dfrac{5}{{16}} + \dfrac{{11}}{{16}}} \right) + 2 + \dfrac{{15}}{{17}}\\
= \dfrac{{ - 15}}{{17}}.\dfrac{{16}}{{11}} + 2 + \dfrac{{15}}{{17}}\\
= - \dfrac{{15}}{{17}} + 2 + \dfrac{{15}}{{17}}\\
= 2\\
D = \dfrac{2}{{4.7}} + \dfrac{2}{{7.10}} + ... + \dfrac{2}{{34.37}}\\
= \dfrac{2}{3}.\left( {\dfrac{3}{{4.7}} + \dfrac{3}{{7.10}} + ... + \dfrac{3}{{34.37}}} \right)\\
= \dfrac{2}{3}.\left( {\dfrac{1}{4} - \dfrac{1}{7} + \dfrac{1}{7} - \dfrac{1}{{10}} + ... + \dfrac{1}{{34}} - \dfrac{1}{{37}}} \right)\\
= \dfrac{2}{3}.\left( {\dfrac{1}{4} - \dfrac{1}{{37}}} \right)\\
= \dfrac{2}{3}.\dfrac{{33}}{{148}}\\
= \dfrac{{11}}{{74}}
\end{array}$
