thi xong r hả bae :3

=> :( thi khomg được tốt lém 

Ta có 

\(\widehat{FGE}+x=180^o\)(hai góc kề bù )

\(\Rightarrow x=180^o-40^o=140^o\)

Theo định lý tổng 3 góc trong tam giác 

\(\widehat{E}+\widehat{FGE}+\widehat{EFG}=180^o\\ \Leftrightarrow\widehat{EFG}=180^o-40^o-60^o=80^o\)

Ta có

\(y+\widehat{EFG}=180^o\) (hai góc kề bù)

\(\Rightarrow y=180^o-80^o=100^o\)

A

Giả sử \(\left(d'\right):y=ax+b\)

\(\left(d'\right)//\left(d\right)\)

\(\Rightarrow\) phương trình : \(\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b\ne-1\end{matrix}\right.\)

\(\Rightarrow y=-x+b\)

cắt (P)tại diểm có hoành độ =4

\(A=\left(\dfrac{x+8}{\left(\sqrt{x}\right)^3+8}-\dfrac{2}{x-2\sqrt{x}+4}\right):\dfrac{1}{\sqrt{x}-1}\\ =\dfrac{x+8-2.\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}\times\dfrac{\sqrt{x}-1}{1}\\ =\dfrac{x+8-2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}.\dfrac{\sqrt{x}-1}{1}\\ =\dfrac{x-2\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}.\left(\sqrt{x}-1\right)\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

come back roi a

=> bận thi á :< 

1- My house is bigger than your house

2- is more beautiful than

3- the most interesting

4- longer than 

5- the most dangerous

6- better than 

7- more expensive than 

8- the richest 

9- worse than

10- the cleverest 

1- was

2- weren't

3- were