Trên  cùng nửa mặt phẳng có bờ chứa tia Ox có :

\(\widehat{xom}< \widehat{xon}\) ( 40⁰ < 80⁰)

=> Om là tia nằm giữa 2 tia Ox và On  (1)

nên \(\widehat{xom}+\widehat{mon}=\widehat{xon}\) 

\(\Rightarrow\widehat{mon}=\widehat{xon}-\widehat{xom}=80^0-40^0=40^0\)

Vậy \(\widehat{xom}=\widehat{mon}=40^0\) (2)

từ (1) và (2)=> Om là tia phân giác của góc xOn 

b, ( câu này mình không chăc lắm )

góc kề bù với góc xOn là : nOy 

Vì Xon và nOy  là 2 góc kề bù nên 

\(\widehat{xon}+\widehat{noy}=180^0\) 

\(\Rightarrow\widehat{noy}=180^0-\widehat{xon}=180^0-80^0\) \(=100^0\)

-25*72+25*21-49*25

=(-25).(72-21+49)

=(-25).100

=(-2500)

bn nhớ k nha

\(A=1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}\)

\(2A=2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}\)

\(\Rightarrow A=2A-A=2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}-\left(1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}\right)\)

\(=2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2019}}-1-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2020}}\)

\(=1+\frac{1}{2}-\frac{1}{2^{2020}}=\frac{2^{2020}+2^{2019}-1}{2^{2020}}\)

\(A=\frac{7}{4}.\)\(\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)

\(A=\frac{7}{4}.\)\(\left[3333.\left(\frac{1}{1212}+\frac{1}{2020}+\frac{1}{3030}+\frac{1}{4242}\right)\right]\)

\(A=\frac{7}{4}.\)\(\left[3333.\left(\frac{1}{12.101}+\frac{1}{20.101}+\frac{1}{30.101}+\frac{1}{42.101}\right)\right]\)

\(A=\frac{7}{4}.\)\(\left[3333.\frac{1}{101}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]\)

\(A=\frac{7}{4}.\)\(33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(A=33.\left[\frac{7}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\right]\)

\(A=33.\left[\frac{7}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)

\(A=33.\left(\frac{7}{4}.\frac{4}{21}\right)\)

\(A=33.\frac{1}{3}\)

\(A=11\)

 Vậy \(A=11\)