\(SA\perp\left(ABCD\right)\Rightarrow\left\{{}\begin{matrix}SA\perp AB\\SA\perp AD\end{matrix}\right.\) \(\Rightarrow\) các tam giác SAB và SAD vuông tại A

\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp BC\\BC\perp AB\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\Rightarrow BC\perp SB\)

\(\Rightarrow\Delta SBC\) vuông tại B

\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp CD\\CD\perp AD\end{matrix}\right.\) \(\Rightarrow CD\perp\left(SAD\right)\Rightarrow CD\perp SD\)

\(\Rightarrow\Delta SCD\) vuông tại D

b.

\(\left\{{}\begin{matrix}BC\perp\left(SAB\right)\Rightarrow BC\perp AH\\AH\perp SB\end{matrix}\right.\) \(\Rightarrow AH\perp\left(SBC\right)\Rightarrow AH\perp SC\) (1)

\(\left\{{}\begin{matrix}CD\perp\left(SAD\right)\Rightarrow CD\perp AK\\AK\perp SD\end{matrix}\right.\) \(\Rightarrow AK\perp\left(SCD\right)\Rightarrow AK\perp SC\) (2)

(1);(2) \(\Rightarrow SC\perp\left(AHK\right)\)

B nhé

\(g'\left(x\right)=3.f'\left(3x\right)+9=0\Rightarrow f'\left(3x\right)=-3\Rightarrow\left[{}\begin{matrix}3x=-1\\3x=0\\3x=1\\3x=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=0\\x=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(\Rightarrow\) Trên \(\left[-\dfrac{1}{3};\dfrac{1}{3}\right]\) hàm \(g\left(x\right)\) đạt cực đại tại \(x=0\) và cực tiểu tại \(x=-\dfrac{1}{3};\dfrac{1}{3}\)

\(\Rightarrow g\left(x\right)_{max}=g\left(0\right)=f\left(0\right)\)

22,play tennis better than his

24,cheaper than this one

26,stronger than the girls

28 drives more carefully than his

12. not studying hard,.....

14, living in a poor family,......

16,difficult  test ,....

18,good manner,...

20,

Bài 3: 

a: =>5x-30=25

hay x=11

Bài 3: 

a: =>5x-30=25

hay x=11

no good thơ lục bát

c) C = \(\dfrac{4}{\sqrt{3}+1} - \dfrac{5}{\sqrt{3}-2} + \dfrac{6}{\sqrt{3}-3}\)
⇔ C = \(\dfrac{4(\sqrt{3}-1)}{2} - \dfrac{5(\sqrt{3}-2)}{-1} - \dfrac{6(\sqrt{3}+3)}{-6}\)
⇔ C = \(2\sqrt{3} -2 + 5\sqrt{3} + 10 - \sqrt{3} - 3\)
⇔ C = \(6\sqrt{3} + 5\)