O y B N M x A

\(a)\)

\(\widehat{AOM}=2-\widehat{BOM}=2-90^o\)

\(\widehat{BON}=2-\widehat{AON}=2-90^o\)

\(\Rightarrow\widehat{AOM}=\widehat{BON}\)

\(b)\)

Ox là tia phân giác của \(\widehat{AOM}\Rightarrow\widehat{xOA}=\frac{1}{2}\widehat{AOM}=\frac{1}{2}\left(2-90^o\right)\)

Oy là tia phân giác của \(\widehat{BON}\Rightarrow\widehat{yOB}=\frac{1}{2}\widehat{BON}=\frac{1}{2}\left(2-90^o\right)\)

\(\Rightarrow\widehat{xOy}=\widehat{AOB}-2\frac{1}{2}\left(2-90^o\right)=2-2+90^o=90^o\)

Vậy \(Ox\perp Oy\)

 AOB < BOy => OA nằm giữa Ox và OB

=> xOA + AOB = BOx

xOA = BOx - AOB = 90°  - 60° = 30° 

aOx < aOy => OB nằm giữa OA và Oy

=> AOB + BOy = AOy

BOy = AOy - AOB = 90°  - 60°  = 30° 

=> AOx = BOy

Cre : lazi

bằng boy

mình ko viết dài dòng nhé vì olm chỉ bảo ghi đấp án thui

\(\frac{31}{2}\times\frac{32}{2}\times\frac{33}{2}\times...\times\frac{60}{2}=\frac{1\times2\times...\times60}{2^{30}\times\left(1\times2\times...\times30\right)}\)

\(=\frac{1\times2\times...\times60}{\left(2\times1\right)\times\left(2\times2\right)\times...\times\left(2\times30\right)}=\frac{\left(2\times4\times...\times60\right)\times\left(1\times3\times...\times59\right)}{2\times4\times...\times60}=1\times3\times...\times59\)

a) \(\frac{\left(-3\right)^{10}.15^5}{25^3.\left(-9\right)^7}=\frac{\left(-3\right)^{10}.3^5.5^5}{\left(5^2\right)^3.\left(-3^2\right)^7}=\frac{\left(-3\right)^{10}.3^5.5^5}{5^6.\left(-3\right)^{14}}=\frac{3^5}{5.\left(-3\right)^4}=\frac{-3}{5}\)

\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)

\(=\frac{100}{2}\)

\(=50\)