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)

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

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

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

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

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

Why did you stay at home?

Who are they waiting for?

What is she doing at the moment?

How did she go to Ha Noi? 

When does your father often watch TV?

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.

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