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Xét tứ giác MNPQ:

Ta có: \(\widehat{M}+\widehat{N}+\widehat{P}+\widehat{Q}=360^o.\)

Mà \(\left\{{}\begin{matrix}\widehat{N}=2\widehat{M}.\\\widehat{P}=\widehat{M}+10^o.\\\widehat{Q}=2\widehat{N}=2.2\widehat{M}=4\widehat{M}.\end{matrix}\right.\)

\(\Rightarrow\widehat{M}+2\widehat{M}+\widehat{M}+10^o+4\widehat{M}=360^o.\\ \Leftrightarrow8\text{​​}\text{​​}\widehat{M}=350^o.\\ \Leftrightarrow\widehat{M}=43,75^o.\)

\(\Rightarrow\widehat{N}=87,5^o.\\ \widehat{P}=53,75^o.\\ \widehat{Q}=175^o.\)

\(\dfrac{11}{9}+\dfrac{7}{6}\times\left(2-\dfrac{1}{3}\right)=\dfrac{11}{9}+\dfrac{7}{6}\times\dfrac{5}{3}=\dfrac{11}{9}+\dfrac{35}{18}=\dfrac{19}{6}.\)

\(A=\dfrac{1+2+3+...+200}{6+8+10+...+34}.\)

- Số số hạng ở tử số: 

   \(\left(200-1\right):1+1=200\) (số).

- Tổng các số ở tử số:

   \(\dfrac{\left(200+1\right).200}{2}=20100.\)

- Số số hạng ở tử số: 

   \(\left(34-6\right):2+1=15\) (số).

- Tổng các số ở tử số: 

   \(\dfrac{\left(34+6\right).15}{2}=300.\)

\(\rightarrow A=\dfrac{1+2+3+...+200}{6+8+10+...+34}=\dfrac{20100}{300}=67.\)

\(e,\left(x-2\right)\left(x+2\right).3xy.\\ =\left(x^2-4\right).3xy.\\ =3x^3y-12xy.\\ g,\left(x+y\right)\left(x-y\right)\left(x+1\right).\\ =\left(x^2-y^2\right)\left(x+1\right).\\ =x^3+x^2-xy^2-y^2.\)