C

Bài 7:

Ta thấy: $\widehat{xOy}+\widehat{yOx'}=\widehat{xOx'}=180^0$

$\widehat{xOy}-\widehat{yOx'}=30^0$

$\Rightarrow \widehat{yOx'}=\frac{180^0-30^0}{2}=75^0$

$\widehat{xOy'}=\widehat{yOx'}=75^0$ (hai góc đối đỉnh)

Bài 8:

$\widehat{AOC}+\widehat{BOD}=140^0$

$\widehat{AOC}=\widehat{BOD}$ (hai góc đối đỉnh)

$\Rightarrow \widehat{AOC}=\widehat{BOD}=\frac{140^0}{2}=70^0$

$\widehat{COB}=180^0-\widehat{AOC}=180^0-70^0=110^0$

$\widehat{DOA}=\widehat{COB}=110^0$ (hai góc đối đỉnh)

giúp mà nha mn

bài 1

a. 15.17-15.16=15.(17-16)=15.1=15

b.73.2²+2².27=2².(73+27)=2².100=4.100=400

c.14:\(\left\{300:\left[500-\left(105+35.7\right)\right]\right\}\)

=14:\(\left\{300:\left[500-\left(105+245\right)\right]\right\}\)

=14:\(\left\{300:\left[500-350\right]\right\}\)=14:\(\left\{300:150\right\}\)=14:2=7

d.14.51+14.6-47.14=14.(51+6-47)=14.10=140

e.3.5²-16:2²=3.25-16:4=75-4=71

f.306:\(\left[119-\left(23-6\right)\right]\)=306:\(\left[119-17\right]\)=306:102=3

\(\left\{{}\begin{matrix}\widehat{OAB}=\widehat{ODC}\left(AB\text{//}CD\right)\\\widehat{OBA}=\widehat{OCD}\left(AB\text{//}CD\right)\\AB=CD\end{matrix}\right.\Rightarrow\Delta OAB=\Delta OCD\left(g.c.g\right)\\ \Rightarrow OA=OD;OB=OC\)

Đặt M (x) = 0

Tức là 2x -6 = 0

2x = 6

x = 3

Vậy nghiệm của đa thức M(x) = 3

Câu 36: -6(x+7)=96

=>x+7=-16

=>x=-16-7

=>x=-23

=>Chọn D

Câu 35: B

Câu 34: -5-x=-11

=>x+5=11

=>x=11-5

=>x=6

=>Chọn A

câu 33: B

câu 32: A

Câu 31: C

Câu 30: A

Câu 29: B

Câu 28: A

Câu 22: A

Câu 23: A

Câu 24: B

Câu 25: A

Câu 26: C

Câu 27: D

Câu 10: D

Câu 11: B

Câu 12: A

Câu 13:A

Câu 15: D

Câu 16: C

Câu 17: C

câu 18: B

Câu 19: A

Câu 20: D

Câu 21: B

B

Ex2

2 It is a library

3 It is a railway station

4 It is a gym

5 It is a square

6 It is an art gallery

Ex3

1 colder

2 more dangerous

3 more important

4 narrower

5 faster

Ex4

1 than he used to be

2 than it was when I bought it

3 she was a child

4 than I was before

5 than they were ten years ago

\(b,\text{PT hoành độ giao điểm: }-2x=3x+3\\ \Leftrightarrow x=-\dfrac{3}{5}\Leftrightarrow y=\dfrac{6}{5}\\ \Leftrightarrow A\left(-\dfrac{3}{5};\dfrac{6}{5}\right)\)