b) Ta có \(\text{∠}yBC=360\text{ ° }-\left(\text{∠}yBA+\text{∠}ABC\right)=360\text{ ° }-\left(120\text{ ° }+90\text{ ° }\right)=150\text{ ° }\)

Mặt khác, ta có \(\text{∠}BCz=150\text{ ° }\)

\(\Rightarrow\text{∠}yBC=\text{∠}BCz\) mà hai góc này ở vị trí so le trong 

\(\Rightarrow By//Cz\) ( đpcm )

mn xa lánh em quá hic:((((((

giúp em với mai em nộp bài đó ạ;-;

a. \(\widehat{DAB}=\widehat{ABC}=\widehat{BCE}=90^0\)

\(\widehat{ABD}=180^0-\widehat{ABC}-\widehat{EBC}=180^0-60^0-\left(180^0-\widehat{BCE}-\widehat{CEB}\right)=180^0-60^0-\left(180^0-60-\widehat{CEB}\right)=\widehat{CEB}\)\(\Rightarrow\)△ABD∼△CEB (g-g).

\(\Rightarrow\dfrac{AD}{CB}=\dfrac{AB}{CE}\Rightarrow AD.CE=CB.AB\Rightarrow AD.CE=a^2\) không đổi

b. \(\widehat{CAD}=\widehat{BAD}+\widehat{BAC}=60^0+60^0=\widehat{BCE}+\widehat{ACB}=\widehat{ACE}\)

 \(\dfrac{AD}{CB}=\dfrac{AB}{CE}\Rightarrow\dfrac{AD}{AC}=\dfrac{AC}{CE}\)

\(\Rightarrow\)△ACD∼△CEA (c-g-c) 

\(\Rightarrow\left\{{}\begin{matrix}\widehat{ACD}=\widehat{CEA}\\\dfrac{CE}{AC}=\dfrac{EA}{CD}\end{matrix}\right.\)

\(\Rightarrow\)△ACK∼△AEC (g-g).

\(\Rightarrow\dfrac{CK}{EC}=\dfrac{AK}{AC}\Rightarrow\dfrac{CE}{AC}=\dfrac{CK}{AK}\)

\(\Rightarrow\dfrac{AE}{CD}=\dfrac{CK}{AK}\Rightarrow AE.AK=CD.CK\)

Bài 4: 

Nhóm 1: x;1/3x; 8x

Nhóm 2: \(x^2;5x^2;-3x^2\)

câu 1 cậu tự làm nha

UKM

bạn tự làm nhé!

1 D

2 A

3 B

4 A

5 C

6 D

7 A

8 B

9 C

10 D

11 A

12 B

13 B

14 B

15 A

16 B

17 B

18 A

19 B

20 D

21 D

22 C

23 D

24 A

25 C

26 A

27 C

28 B

29 A

30 D

31 A

32 B

33 A

34 B

35 A

36 A

37 A

38 B

39 D

40 A

41 A

42 B

43 B

44 B

45 D

46 A

47 A

48 B

49 D

50 D

1 D

2 A

3 B

4 A

5 C

6 D

7 A

8 B

9 C

10 D

11 A

12 B

13 B

14 B

15 A

16 B

17 B

18 A

19 B

20 D

21 D

22 C

23 D

24 A

25 C

26 A

\(A=\dfrac{1}{200\cdot199}-\dfrac{1}{199\cdot198}-\dfrac{1}{198\cdot197}-...-\dfrac{1}{3\cdot2}-\dfrac{1}{2\cdot1}\)

\(=\dfrac{1}{199}-\dfrac{1}{200}-\left(1-\dfrac{1}{2}-...+\dfrac{1}{198}-\dfrac{1}{199}\right)\)

\(=\dfrac{1}{199}-\dfrac{1}{200}-1+\dfrac{1}{199}\)

\(=\dfrac{-39599}{39800}\)

Bn chụp thiếu rồi!

\(\dfrac{1}{200.199}-\dfrac{1}{199.198}-\dfrac{1}{198.197}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=-\left(\dfrac{1}{200.199}+\dfrac{1}{199.198}+\dfrac{1}{198.197}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{197.198}+\dfrac{1}{198.199}+\dfrac{1}{199.200}\right)=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{197}-\dfrac{1}{198}+\dfrac{1}{198}-\dfrac{1}{199}+\dfrac{1}{199}-\dfrac{1}{200}\right)=-\left(1-\dfrac{1}{200}\right)=\dfrac{1}{200}-1=-\dfrac{199}{200}\)

Lời giải:

Lời giải:

$372802=300000+70000+2000+800+2$

$430279=400000+30000+200+70+9$
$920300=900000+20000+300$

$704753=700000+4000+700+50+3$

a: Ta có: \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{8}\\x=-\dfrac{3}{8}\end{matrix}\right.\)

b: Ta có: \(\left(\dfrac{1}{2}\right)^x=\dfrac{1}{32}\)

nên x=5