\(A=\dfrac{31\cdot\left(31^{12}-1\right)}{31\left(31^{13}+1\right)}=\dfrac{31^{13}+1-32}{31\left(31^{13}+1\right)}=\dfrac{1}{31}-\dfrac{32}{31^{14}+31}\)

\(B=\dfrac{31\left(31^{13}-1\right)}{31\left(31^{14}+1\right)}=\dfrac{1}{31}-\dfrac{32}{31^{15}+31}\)

Dễ thấy \(31^{14}+31< 31^{15}+31\Rightarrow\dfrac{32}{31^{14}+31}>\dfrac{32}{31^{15}+31}\\ \Rightarrow\dfrac{1}{31}-\dfrac{32}{31^{14}+31}< \dfrac{1}{31}-\dfrac{32}{31^{15}+31}\)

Vậy A < B

5A

1B

3C

4D

\(#TyHM\)

1 do you do

2 had done - went

3 went - had read

4 will attend

5 hadn't worn

6 to be

7 weren't sleeping - were playing

8 to be

9 had lived - moved

10 locking

11 had work - retired 

12 told - had learned

13 won't call

14 had met

do you do

had done-went

went- had read

will attend

hadn't worn

to be

a: Xét tứ giác AEHF có 

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

Do đó: AEHF là hình chữ nhật

a: AI=8cm

=>AB=16cm

b: Xét ΔMAO và ΔMBO có 

OA=OB

\(\widehat{MOA}=\widehat{MOB}\)

OM chung

Do đó: ΔMAO=ΔMBO

Suy ra: \(\widehat{MAO}=\widehat{MBO}=90^0\)

hay MB là tiếp tuyến của (O)

Câu 22: A

Câu 24: C

Câu 21: A

Bài 1: 

Ta có: BC=BH+HC

nên BC=9+16

hay BC=25cm

Áp dụng hệ thức lượng trong tam giác vuông vào ΔABC vuông tại A có AH là đường cao ứng với cạnh huyền BC, ta được:

\(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AB^2=9\cdot25\\AC^2=16\cdot25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AB=15cm\\AC=20cm\end{matrix}\right.\)

1 were - would you play

2 weren't studying - would have 

3 had taken - wouldn't have got

4 would you go - could

5 will you give - is

6 recycle - won't be

7 had heard - wouldn't have gone

8 would you buy - had

9 don't hurry - will miss

10 had phoned - would have given

11 were - wouldn't eat

12 will go - rains

13 had known - would have sent

14 won't feel - swims

15 hadn't freezed - would have gone