Đặt 8t=2x

\(d\left(8t\right)=2dx\Rightarrow\frac{d\left(8t\right)}{2}=dx\)

Đổi cận x=0 t=0       x=8 t=2

a , ta có:AE//CF (vì cùng vuông góc vsBD)

=> góc FCO= góc EAO (vì so le trong )

      OA = OC (theo t/c hình bh )

xét 2 tam giác vuông OAE và OCF có:

           góc FOC = góc EAO ( cm trên )

            OA = OC (cmt)

   =>tg OAE = tg OCF (cạnh huyền - góc nhọn )

   =>OE = OF ( 2 cạnh tương ứng )

 b. ta có : AE// CF ( theo a ) (1)

               AE = CF ( vì tg OAE= tg OCF ( theo a )) (2)

 từ (1) và (2) => AECF là hbh

 ( hi vọng đúng !!)

tam giác ABC vuông tại A có AT là đường cao 

Áp dụng định lí Py ta go ta có : \(AB^2+AC^2=BC^2\Rightarrow25-AB^2=AC^2\)(1) 

* Theo hệ thức : \(\frac{1}{AB^2}+\frac{1}{AC^2}=\frac{1}{AT^2}\Rightarrow\frac{1}{4}=\frac{1}{AB^2}+\frac{1}{25-AB^2}\)( theo 1 ) 

\(\Rightarrow AB=2\sqrt{5};\sqrt{5}\)

TH1 : \(25-\left(2\sqrt{5}\right)^2=AC\Rightarrow AC=\sqrt{5}\)

TH2 : \(25-\left(\sqrt{5}\right)^2=AC\Rightarrow AC=2\sqrt{5}\)

Gọi BH là z ( z>0), thì HC là 5-z

ΔABC vuông tại A có:

AH.BC=BH.HC (định lý 3)

⇔ 2² = z(5-z)

⇔ z² - 5z + 4 = 0

⇔ z(z-1) - 4(z-1) = 0

⇔(z-4)(z-1)=0

⇔\(\left[{}\begin{matrix}z-4=0\\z-1=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}z=4\left(nhận\right)\\z=1\left(nhận\right)\end{matrix}\right.\)

TH1:Nếu z=4

ΔABC vuông tại A có:

x²=BC.BH ( định lý 1)

⇔ x²= 5.4

⇔ x²= 20

⇒x=\(2\sqrt{5}\)

ta có: y²= BC.HC ( định lý 1)

Chứng minh tương tự như trên ta được

y= \(\sqrt{5}\)

TH2: Nếu z=1

Chứng minh tương tự như TH1 ta được:

x=\(\sqrt{5}\)

y= \(2\sqrt{5}\)

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2. It is the second nicest cavern in Viet Nam, after Huong rich Cavern.

3. Bich Dong Pagoda was built in Le Dynasty in Bich Dong Cavern which/whose name means “Green Pearl Cavern”.

4. Tam Coc Cavern consists of Ca cave, Hai cave and Ba cave, all of which offer both beauty and mystery to tourists.

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2. A French architect designed the palace.

3. They have paved the road in front of my house.

4. You can see that they haven't washed the dishes.

5. They will not increase our salaries this year.

6. Didn't they build that theatre two years ago?

7. They may discuss the problem again.

8. They have offered my brother a well – paid job.

9. People reported that the war started again in South American.

10. He told me that his football team had played well last season.

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a, xét từ giác AMNC có 
$={90}^{\circ }$ (Ac là tiếp tuyến của (O) , $\stackrel{}{}$

${}^{}={90}^{\circ }$ (MN vuông góc với CD) => $\stackrel{}{}$\(\widehat{CAM}+\widehat{CNM}\)=180

=> AMNC nội tiếp

Xét tứ giác BMND có $\stackrel{ˆ}{}$=90 ( BD là tiếp tuyến của (O) , \(\widehat{CND}\)=90 ( MN vuông góc với CD)

=> \(\widehat{MND}+\widehat{NAC}\)=180

=> Tứ giác BDMN nội tiếp

b, Ta có \(\widehat{CMN}=\widehat{NAC}\) (cùng chắn CN)

=> ${\mathrm{\backslash (\backslash widehat\{CMN\}\backslash )}}^{}$=$\frac{1}{2}$ cung AN(1)

Ta cũng có\(\widehat{NMD}+\widehat{NMD}\) (cùng chắn cung ND)

\(\widehat{NMD}\)=$\frac{1}{2}$ cung NB(2)

Từ (1) và (2) => \(\widehat{CMD}+\widehat{NMD}\)= $\frac{1}{2}$ (cung AN + cung NB) 

=> \(\widehat{CMD}\)= $\frac{1}{2}$ cung AB = $\frac{180}{2}$=90

=> tam giác CMD vuông tại M

Vì NMBD nội tiếp => \(\widehat{NDM}+\widehat{NBM}\) ( góc nội tiếp cùng chắn cung AM) 

Mà \(\widehat{MCD}+\widehat{NBM}\)=90

=> \(\widehat{MCD}+\widehat{NBM}\)=90 (1)

Mặt khác \(\widehat{...

Add: Tr Ph Thảo (hpthaoo)