Mặt trời.

Chắc vậy hoặc không.

sai rồi

là con tem

a) Vì \(\widehat{xOz}-\widehat{yOz}=4\widehat{yOz}\) nên \(\widehat{xOz}=5\widehat{yOz}\)

Mà \(\widehat{xOz},\widehat{yOz}\) kề bù 

\(\Rightarrow\widehat{xOz}+\widehat{yOz}=180^o\)

\(\Rightarrow\widehat{xOz}=180^o:\left(5+1\right).5=150^o\)

\(\Rightarrow\widehat{yOz}=180^o-150^o=30^o\)

Vậy \(\widehat{yOz}=30^o,\widehat{xOz}=150^o\).

b) Trên cùng 1 nửa mặt phẳng có bờ chứa tia Ox, vẽ tia Om sao cho \(\widehat{xOm}=75^o\)

\(\Rightarrow\widehat{xOz}>\widehat{xOm}\left(150^o>30^o\right)\) nên tia Om nằm giữa 2 tia Ox, Oz (1)

Ta có: \(\widehat{xOm}+\widehat{mOz}=\widehat{xOz}\)

\(75^o+\widehat{zOm}=150^o\)

\(\Rightarrow\widehat{zOm}=150^o-75^o=75^o\)

\(\Rightarrow\widehat{xOm}=\widehat{mOz}=\frac{\widehat{xOz}}{2}\) (2)

Từ (1), (2) suy ra Om là tia phân giác của góc xOz.

Vậy Om là tia phân giác của góc xOz.

c) Vì On là tia phân giác của góc yOz 

\(\Rightarrow\)Tia On nằm giữa 2 tia Oy, Oz và \(\widehat{yOn}=\widehat{nOz}=\frac{\widehat{yOz}}{2}=30^o:2=15^o\)

Mà Oz nằm giữa 2 tia Om, On nên ta có:

\(\widehat{nOz}+\widehat{zOm}=\widehat{mOn}\)

\(15^o+75^o=\widehat{mOn}\)

\(\widehat{mOn}=90^o\)   (đpcm)

vẽ hình đc ko bạn

Tác dụng : nhân hóa

Nhân hóa???

\(S=2^2+4^2+6^2+...+20^2\)

\(S=\left(2.1\right)^2+\left(2.2\right)^2+\left(2.3\right)^2+...+\left(2.10\right)^2\)

\(S=2^2\left(1^2+2^2+3^2+...+10^2\right)\)

\(S=4.385=1540\)

Ta có S = 2²(1² + 2² + 3² + .... + 10²)

Đặt P = 1² + 2² + 3² + .... + 10²

= 1.1 + 2.2  + 3 .3 + .... + 10.10

= 1.(2 - 1) + 2.(3 - 1) + 3(4 - 1) + .... + 10(11 - 1)

= 1.2 + 2.3 + 3.4 + ... + 10.11 - (1 + 2 + 3 +... + 10)

= 1.2 + 2.3 + 3.4 + ... + 10.11 - 55

Đặt Q = 1.2 + 2.3 + 3.4 + ... + 10.11 

=> 3Q = 1.2.3 + 2.3.3 + 3.4.3 + ... + 10.11.3

=> 3Q = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + .... + 10.11(12 - 9)

=> 3Q = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 10.11.12 - 9.10.11

=> 3Q = 10.11.12

=> Q = 440

Khi đó P = 440 - 55 = 385

Khi đó S = 2².385 = 4.385 = 1540

1. Ngày mai

2. Ngọn lửa 

3. Bí mật

4. Que diêm

5. Đá lên trời 

6. Mưa

7. Cổ đại

8. Ao , sông , biển

Thank you nha!!!!!!!!!

gọi ah là dường cao tam giác abc

Sabc= \(\frac{ah\cdot8}{2}\)

S tam giác mới=\(\frac{ah.\left(8+5\right)}{2}\)=\(\frac{13.ah}{2}\)

mà nếu cạnh bc tăng thêm 5 cm thi diện tích tam giác abc tăng thêm 15 cm2

=>\(\frac{13ah}{2}\)-\(\frac{8ah}{2}\)=15 

=>\(\frac{5ah}{2}\)=15 

\(\frac{ah}{2}\)=3

ah=6

=> Sabc=\(\frac{6\cdot8}{2}\)=24\(cm^2\)

kho hieu qua

bn ơi cho mk hỏi bn lm tiếng anh hay toán mà chủ đề là tiếng anh mà bài lại là toán vậy alo????

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I D C A B H M K

Bài làm:

Từ A kẻ đường vuông góc với DM cắt DM tại K

Mà AB // KH và AK // BH ( vì cùng vuông góc với DM ) ; góc AKH = 90 độ

=> ABHK là hình chữ nhật

=> AB = HK (1)

Δ ADK = Δ BCH ( c.h-g.n)

=> DK = HC

Mà DH = HM <=> DK + KH = HC + CM

=> KH = CM (2)

Từ (1) và (2) => AB = CM, mà AB // CM

=> Tứ giác ABMC là hình bình hành

=> BM = AC