Hứa Tick

Kẻ CD//AB thì CD//MN

Do đó \(\widehat{ACD}=\widehat{CAB}=41^0;\widehat{MCD}=\widehat{CMN}=54^0\) (so le trong)

Vậy \(\widehat{ACM}=\widehat{ACD}+\widehat{DCM}=41^0+54^0=95^0\)

Gọi d=ƯC(2n+7;5n+17)

\(\Rightarrow\left\{{}\begin{matrix}2n+7⋮d\\5n+17⋮d\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}10n+35⋮d\\10n+34⋮d\end{matrix}\right.\)

\(\Rightarrow1⋮d\Rightarrow d=1\)

\(\Rightarrow\) phân số \(\dfrac{2n+7}{5n+17}\) tối giản

\(2A=\dfrac{4}{1.5}+\dfrac{6}{5.11}+\dfrac{8}{11.19}+\dfrac{10}{19.29}+\dfrac{12}{29.41}\)

\(2A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{41}\)

\(2A=1-\dfrac{1}{41}=\dfrac{40}{41}\)

\(A=\dfrac{40}{41}:2=\dfrac{20}{41}\)

\(3B=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}\)

\(3B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}\)

\(3B=1-\dfrac{1}{31}=\dfrac{30}{31}\)

\(B=\dfrac{30}{31}:3=\dfrac{10}{31}\)

bài đâu bạn

I

1 cartoon

2 remote control

3 weather man

4 national television

5 game show

6 comedy

7 TV schedule

8 news programme

II

1 newsreader

2 Remote control

3 game show

4 TV schedule

5 comedy

II

1 B

2 C

3 A

4 D

5 B

IV

Cột 1: think, north, earth, throw, thousand, truth, teeth, booth

Cột 2 là các từ còn lại

V

1 D

2 A

3 B

4 C

5 B

6 D

7 C

VI

1 B

2 A

3 B

4 A

5 C

`19dm^3=19 : 1000= 0,019 m^3`

`5m^3 19dm^3=5+0,019=5,019m^3`

`5m^3 19dm^3 =5+19:1000=5,019m^3`

Lời giải:

$b=a+1=5+1=6$. Khi đó:

$(a+b)^2-(b-a)^3+2021=(5+6)^2-(6-5)^3+2021$

$=11^2-1^3+2021=121-1+2021=2141$

đúng không đó ạ

Xét tứ giác MECF có 

ME//CF

MF//EC

Do đó: MECF là hình bình hành

Suy ra: ME=CF, MF=EC

ME+MF=CF+EC ko đổi

30,27

Cảm ơn Bn Thịnh nhaaa

30,27