c, 18/-5 . -15/20

= 18.(-15)/(-5).20

= 270/100 = 27/10

d, 115/30 : - 5/6

= 115/30 . 6/-5

= 115.6/30.(-5)

= 690/-15

= -46

HỌC TỐT!

Trả lời:

= -46 nhé bn\

~HT~

Kick cho mik nha:33

Bài 1: - \(\dfrac{5}{7}\) x \(\dfrac{31}{33}\) + \(\dfrac{-5}{7}\) x \(\dfrac{2}{33}\) + 2\(\dfrac{5}{7}\)

         = - \(\dfrac{5}{7}\) \(\times\) ( \(\dfrac{31}{33}\) + \(\dfrac{2}{33}\)) + 2 + \(\dfrac{5}{7}\)

         =  - \(\dfrac{5}{7}\)  + 2 + \(\dfrac{5}{7}\)

          = 2

2, \(\dfrac{3}{14}\): \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\): \(\dfrac{1}{28}\) - 8

   = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) : \(\dfrac{1}{28}\) - 8

   =  \(\dfrac{2}{7}\) x 28 - 8

   = 8 - 8

    = 0 

3, \(\dfrac{37}{43}\) . \(\dfrac{17}{29}\) - \(\dfrac{21}{41}\).\(\dfrac{1}{2}\) + \(\dfrac{9}{58}\): 1\(\dfrac{6}{37}\) - \(\dfrac{6}{29}\): 1\(\dfrac{20}{21}\)

=    \(\dfrac{37}{43}\). \(\dfrac{17}{29}\) - \(\dfrac{21}{41}\) . \(\dfrac{1}{2}\) + \(\dfrac{9}{58}\).\(\dfrac{37}{43}\)  - \(\dfrac{6}{29}\).\(\dfrac{21}{41}\)

=    (\(\dfrac{37}{43}\).\(\dfrac{17}{29}\) + \(\dfrac{9}{58}\).\(\dfrac{37}{43}\)) - (\(\dfrac{21}{41}\).\(\dfrac{1}{2}\) + \(\dfrac{6}{29}\).\(\dfrac{21}{41}\))

=     \(\dfrac{37}{43}\).( \(\dfrac{17}{29}\) + \(\dfrac{9}{58}\)) - \(\dfrac{21}{41}\).( \(\dfrac{1}{2}\) + \(\dfrac{6}{29}\))

=       \(\dfrac{37}{43}\).\(\dfrac{43}{58}\) - \(\dfrac{21}{41}\).\(\dfrac{41}{58}\)

=         \(\dfrac{37}{58}\) - \(\dfrac{21}{58}\)

=          \(\dfrac{16}{58}\)

= \(\dfrac{8}{29}\)

\(\Delta'=4-\left(m-1\right)=5-m\)

để pt có nghiệm kép khi \(5-m=0\Leftrightarrow m=5\)

chọn B 

Phương trình có nghiệm kép khi:

\(\Delta'=4-\left(m-1\right)=0\Leftrightarrow5-m=0\)

\(\Rightarrow m=5\)

Gọi giao điểm AE và BP là F;

Gọi giao điểm QD và AB là H; 

Gọi kéo dài AD cắt BF tại P'     

Dễ cm M là trung điểm AC

Xét \(\Delta OMC\) có QD//CM\(\Rightarrow\dfrac{OD}{OM}=\dfrac{QD}{CM}\)(hệ quả tales)

Tương tự với \(\Delta OAM\) có \(\dfrac{OD}{OM}=\dfrac{DH}{AM}\) 

\(\Rightarrow\dfrac{QD}{CM}=\dfrac{DH}{AM}\)

Mà CM=AM (vì M là tđ AC)

\(\Rightarrow QD=DH\)

Dễ cm P là trung điểm BF

Xét \(\Delta ABP'\) có DH//BP'

\(\Rightarrow\dfrac{DH}{BP'}=\dfrac{AD}{AP'}\)(tales)

Tương tự với \(\Delta AFP'\) có \(\dfrac{QD}{FP'}=\dfrac{AD}{AP'}\)

\(\Rightarrow\dfrac{DH}{BP'}=\dfrac{QD}{FP'}\)

Mà DH=QD (cmt) 

\(\Rightarrow BP'=FP'\)

\(\Rightarrow\)P' là trung điểm BF

\(\Rightarrow P\equiv P'\)

\(\Rightarrow A,D,P\) thẳng hàng

bạn vẽ hình được không?

Bài 1: 

a) Xét ΔMNQ và ΔENQ có 

NM=NE(gt)

\(\widehat{MNQ}=\widehat{ENQ}\)

NQ chung

Do đó: ΔMNQ=ΔENQ(c-g-c)

Suy ra: QM=QE(hai cạnh tương ứng)

Bài 1: 

b) Ta có: ΔQMN=ΔQEN(cmt)

nên \(\widehat{QMN}=\widehat{QEN}\)(hai góc tương ứng)

mà \(\widehat{QMN}=90^0\)(gt)

nên \(\widehat{QEN}=90^0\)

a: Xét tứ giác BHCD có 

CH//BD

BH//CD

Do đó: BHCD là hình bình hành

b: Xét ΔAIC vuông tại I và ΔAKB vuông tại K có 

\(\widehat{A}\) chung

Do đó: ΔAIC\(\sim\)ΔAKB

Suy ra: \(\dfrac{AI}{AK}=\dfrac{AC}{AB}\)

hay \(AI\cdot AB=AK\cdot AC\)

-98.(1-247)+98-247.98

= 98 {(-1+247) +1-247} 

= 98 (-1+1+247-247)

=98. (0+0)

=0

Trả lời:

-98.(1-247)+98-247.98= 98 (-1+247+1-247)=0

9. is/ tells

10. are

11. has

12. isn't

--------

B3

1. is

2. teaches

3. learn

4. comes

5. has

6. goes

7. meets

8. drinks

------------

B4

1. plays

2. drink

3. starts

4. live

5. do

6. speak

7. watches

------------

B5

1. does -> do

2. Do -> Does

3. watch -> watches

4. goes -> go

5. carries -> carry

6. don't -> doesn't

7. plays -> play

8. are -> is

bài nào vs ak

9. is/ tells.

10. are.

11. has.

12. isn't.

B3

1. is.

2. teaches.

3. learn.

4. comes.

5. has.

6. goes.

7. meets.

8. drinks.

B4

1. plays.

2. drink.

3. starts.

4. live.

5. do.

6. speak.

7. watches.

B5

1. does -> do.

2. Do -> Does.

3. watch -> watches.

4. goes -> go.

5. carries -> carry.

6. don't -> doesn't.

7. plays -> play.

8. are -> is.