a: Xét ΔABC vuông tại A có AH là đường cao

nên \(AH^2=HB\cdot HC\)

=>\(AH^2=4\cdot9=36\)

=>AH=6(cm)

BC=BH+CH

=4+9

=13(cm)

Xét ΔABC vuông tại A có AH là đường cao

nên \(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot CB\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AB=\sqrt{4\cdot13}=2\sqrt{13}\left(cm\right)\\AC=\sqrt{9\cdot13}=3\sqrt{13}\left(cm\right)\end{matrix}\right.\)

Xét ΔABC vuông tại A có \(tanABC=\dfrac{AC}{AB}=\dfrac{3}{2}\)

nên \(\widehat{ABC}\simeq56^0\)

b: Xét tứ giác AEHF có

\(\widehat{AEH}=\widehat{AFH}=\widehat{FAE}=90^0\)

=>AEHF là hình chữ nhật

=>AH=EF

Xét ΔHAB vuông tại H có HE là đường cao

nên \(AE\cdot AB=AH^2\)

Xét ΔHAC vuông tại H có HF là đường cao

nên \(AF\cdot AC=AH^2\)

\(AE\cdot AB+AF\cdot AC=AH^2+AH^2=2AH^2=2FE^2\)

<ko hiểu thì ib hỏi mình nha>

a,Thời gian rơi của vật là

\(t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2\cdot125}{10}}=5\left(s\right)\)

Vận tốc của vật ngay trước khi chạm đất

\(v=gt=10\cdot5=50\left(\dfrac{m}{s}\right)\)

b,Quãng đường vật rơi trong giây cuối là

\(\dfrac{1}{2}\cdot10t^2-\dfrac{1}{2}\cdot10\left(t-1\right)^2=25\Rightarrow t=3\left(s\right)\)

độ cao h là

\(s=\dfrac{1}{2}gt^2=\dfrac{1}{2}\cdot10\cdot3^2=45\left(m\right)\)

a. \(t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2\cdot125}{10}}=5\left(s\right)\)

\(v^2=2gh\Rightarrow v=\sqrt{2gh}=\sqrt{2\cdot10\cdot125}=50\left(\dfrac{m}{s}\right)\)

b. Ta có: \(\left\{{}\begin{matrix}h=s=\dfrac{gt^2}{2} \left(1\right)\\h'=s'=\dfrac{g}{2}\left(t-1\right)^2\left(2\right)\end{matrix}\right.\)

\(\Delta h=h-h'=25\left(m\right)\)

\(\Leftrightarrow\dfrac{gt^2}{2}-\dfrac{g}{2}\left(t^2-2t+1\right)=25\)

\(\Leftrightarrow gt-\dfrac{g}{2}=25\)

\(\Rightarrow t=\dfrac{25+5}{10}=3\left(s\right)\left(3\right)\)

\(Thay\left(3\right)in\left(1\right):h=\dfrac{10\cdot3^2}{2}=45\left(m\right)\)

\(VT=\sqrt{\dfrac{\sqrt{5}}{8\sqrt{5}+3\sqrt{35}}}.\left(3\sqrt{2}+\sqrt{14}\right)\)

\(=\sqrt{\dfrac{\sqrt{5}}{8\sqrt{5}+3\sqrt{5}.\sqrt{7}}}.\left(3\sqrt{2}+\sqrt{2}.\sqrt{7}\right)\)

\(=\sqrt{\dfrac{\sqrt{5}}{\sqrt{5}\left(8+3\sqrt{7}\right)}}.\left[\sqrt{2}\left(3+\sqrt{7}\right)\right]\)

\(=\sqrt{\dfrac{1}{8+3\sqrt{7}}}.\left[\sqrt{2}\left(3+\sqrt{7}\right)\right]\)

\(=\dfrac{\sqrt{2}\left(3+\sqrt{7}\right)}{\sqrt{8+3\sqrt{7}}}\)

\(=\dfrac{\sqrt{2}.\sqrt{2}\left(3+\sqrt{7}\right)}{\sqrt{2}.\sqrt{8+3\sqrt{7}}}\) (Nhân \(\sqrt{2}\) cả tử và mẫu)

\(=\dfrac{2\left(3+\sqrt{7}\right)}{\sqrt{16+6\sqrt{7}}}\)

\(=\dfrac{2\left(3+\sqrt{7}\right)}{\sqrt{\left(3+\sqrt{7}\right)^2}}\)

\(=\dfrac{2\left(3+\sqrt{7}\right)}{\left|3+\sqrt{7}\right|}\)

\(=\dfrac{2\left(3+\sqrt{7}\right)}{3+\sqrt{7}}\)

\(=2=VP\left(dpcm\right)\)

...eating lots of food, she is still very slim

...is it from Hanoi to HCM city?

...interested in English

...dance very well 

...the weather was bad, we went to school

...eating lots of food, she is still very slim

did - go

went - didn't enjoy 

take

takes

goes

playing

am wanting

used to play - was

did - ever heard

will - travel

will be taking

will spend

haven't seen

she is interested in reading comic

Tomorrow in this time I wil be lieing on the beach

My car isn't as expensive as his

She has started live in Paris since 2005

A new dam will build near the rive next year

She is interested in reading comic.

Tomorrow in this time I wil be lying on the beach.

My car isn't as expensive as his.

She has started living in Paris since 2005.

A new dam will be built near the rive next year.

1. used to

2. will be taking

3. am watching

4. will go

5. will be washing

6. will be replaced

7. do

8. has visited

9. haven't bought

10. swim

1. had used

2. will be taking

3. will be watching

4. will go

5. will be washing

6. will be replaced

7. do

8. has visited

9. have not bought

10. swim

1. used to

2. will be taking

3. am watching

4. will go

5. will be washing

6. will be replaced

7. do

8. has visited

9. haven't bought

10. swim

1. used to

2. will be taking

3. am watching

4. will go

5. will be washing

6. will be replaced

7. do

8. has visited

9. haven't bought

10. swim