Bài 4:

a: ΔADM vuông tại D

=>\(AD^2+DM^2=AM^2\)

=>\(AM^2=3^2+1^2=9+1=10\)

=>\(AM=\sqrt{10}\left(\operatorname{cm}\right)\)

Xét ΔAMF vuông tại A có AD là đường cao

nên \(AD^2=DM\cdot DF\)

=>\(DF=\frac{3^2}{1}=9\left(\operatorname{cm}\right)\)

ΔADF vuông tại D

=>\(DA^2+DF^2=AF^2\)

=>\(AF^2=3^2+9^2=9+81=90\)

=>\(AF=3\sqrt{10}\left(\operatorname{cm}\right)\)

b: Ta có: \(\hat{BAE}+\hat{DAE}=\hat{BAD}=90^0\)

\(\hat{DAM}+\hat{DAE}=\hat{MAE}=90^0\)

Do đó: \(\hat{BAE}=\hat{DAM}\)

Xét ΔBAE vuông tại B và ΔDAM vuông tại D có

\(\hat{BAE}=\hat{DAM}\)

Do đó: ΔBAE~ΔDAM

=>\(\frac{BE}{DM}=\frac{AE}{AM}\)

=>\(BE\cdot AM=DM\cdot AE\)

c: FC+CD=FD

=>FC=9-6=3(cm)

Xét ΔFCE vuông tại C và ΔFDA vuông tại D có

\(\hat{CFE}\) chung

Do đó: ΔFCE~ΔFDA

=>\(\frac{S_{FCE}}{S_{FDA}}=\left(\frac{FC}{FD}\right)^2=\left(\frac39\right)^2=\frac19\)

Pitago cho tam giác vuông ABM vuông tại M:

\(BM=\sqrt{AB^2-AM^2}=3\)

Hệ thức lượng tam giác vuông ABC: 

\(AB^2=BM.BC\Rightarrow BC=\dfrac{AB^2}{BM}=\dfrac{25}{3}\)

\(\Rightarrow CM=BC-BM=\dfrac{16}{3}\)

\(AC=\sqrt{BC^2-AB^2}=\dfrac{20}{3}\)

I.

B. drives => rides

D. somtimes => by bus (by car,...)

B. hot => cold

B. does => do

B. spend => spending

C. with => in

A. are => is

A. How long => how often

D. later => ago

A. goes => go

V.

I study English on Tuesday

Does your mother drive you to school?

The students are playing right now.

They are playing games on the playground.

Our teacher is short and has long brown hair.

She is wearing a blue T-shirt and jeans.

My friend is an active boy.

I am watching a film with my friend.

My teacher is a smart and funny woman.

Would you like to come to my party?

I'm going to my English club this Sunday.

tách nhỏ câu hỏi ra bạn

c: \(=4\sqrt{5}+4\sqrt{3}-4\sqrt{3}-2\sqrt{5}=2\sqrt{5}\)

d: \(\Leftrightarrow\dfrac{x+14}{86}+1+\dfrac{x+15}{85}+1+\dfrac{x+16}{84}+1+\dfrac{x+17}{83}+1+\dfrac{x+116}{4}-4=0\)

=>x+100=0

hay x=-100

a: \(\Leftrightarrow2\left(x+1\right)^2+3\left(x^2-3x+2\right)=\left(3x-1\right)\left(x-4\right)+56\)

\(\Leftrightarrow2x^2+4x+2+3x^2-9x+6-\left(3x^2-13x+4\right)-56=0\)

\(\Leftrightarrow5x^2-5x-48-3x^2+13x-4=0\)

\(\Leftrightarrow2x^2+8x-52=0\)

\(\Leftrightarrow x^2+4x-13=0\)

\(\Leftrightarrow\left(x+2\right)^2=17\)

hay \(x\in\left\{-\sqrt{17}-2;\sqrt{17}-2\right\}\)