Bài 15:

a) \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{x-1}\right):\left(2-\dfrac{\sqrt{x}-4}{\sqrt{x}-1}\right)\)

\(=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}-4}{\sqrt{x}-1}\right)\)

\(=\left(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{2\left(\sqrt{x}-1\right)-\left(\sqrt{x}-4\right)}{\sqrt{x}-1}\right)\)

\(=\dfrac{\sqrt{x}+1+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+4}{\sqrt{x}-1}\right)\)

\(=\dfrac{\sqrt{x}+2}{x-1}:\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)

\(=\dfrac{\sqrt{x}+2}{x-1}.\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

\(=\dfrac{\sqrt{x}-1}{x-1}\)

\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{1}{\sqrt{x}+1}\)

Bài 16:

a: Ta có: \(Q=\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+3}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{14}{9-x}\right)\cdot\dfrac{\sqrt{x}-3}{2}\)

\(=\dfrac{x-6\sqrt{x}+9+x+6\sqrt{x}+9+14}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2}\)

\(=\dfrac{2x+22}{\sqrt{x}-3}\cdot\dfrac{1}{2}\)

\(=\dfrac{x+11}{\sqrt{x}-3}\)

\(15,A=\dfrac{x-1-4\sqrt{x}+4+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ A=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\\ 16,B=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{2\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\left(\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\\ B=x-\sqrt{x}-2\sqrt{x}-2+2\sqrt{x}+2=x-\sqrt{x}\)

Câu 3: 

a: \(BD=\sqrt{BC^2-DC^2}=4\left(cm\right)\)

b: \(\widehat{A}=180^0-2\cdot70^0=40^0< \widehat{B}\)

nên BC<AC=AB

c: Xét ΔEBC vuông tại E và ΔDCB vuông tại D có

BC chung

\(\widehat{EBC}=\widehat{DCB}\)

Do đó:ΔEBC=ΔDCB

d: Xét ΔOBC có \(\widehat{OBC}=\widehat{OCB}\)

nên ΔOBC cân tại O

Câu 2

a) Thay y = -2 vào biểu thức đã cho ta được:

2.(-2) + 3 = -1

Vậy giá trị của biểu thức đã cho tại y = -2 là -1

b) Thay x = -5 vào biểu thức đã cho ta được:

2.[(-5)² - 5] = 2.(25 - 5) = 2.20 = 40

Vậy giá trị của biểu thức đã cho tại x = -5 là 40

Đăng lại vì ko ai giải 🥺

thôi bọn mềnh cũng chệu boạn nhóe

9 will have to

10 must work

11 will you do - fail

12 will you do

13 were - would buy

14 were - would help

15 had - would travel

16 had - would go

17 would tell - were

18 knew - would speak

19 gets - will go

20 can play - is

21 isn't raining - will have

22 had - would finish

23 wouldn't do - were

24 didn't speak

25 would buy - had

26 weren't - would buy 

27 goes

28 will she go

29 would you do - were

30 were - would help

31 would you buy - were

32 would she do - were

33 will they do - become

34 would Ba go - had

35 had been - would have helped

36 had had - would have bought

37 hadn't bought - would have bought

38 would have been - hadn't rained

39 wouldn't have done

40 would have bought

41 would you do

42 would you have done

43 would you have done

44 would you have bought

45 has

46 would have come

III.

1 I will visit you

2 I would travel around the world

3 we will fail the final term test

4 you should take care of her

5 I would run away

6 we will go on a picnic

7 I would have got a good mark

8 he wouldn't be so fat now

9 I would kill it 

hình 1
biểu đạt tại B
phương ngang
chiều: từ trái sang phải
độ lớn: 30N ứng vs ba đoạn, mỗi đoạn 10N
hình 2
biểu đạt tại D
phương dọc 
chiều: trên xuống dưới, trái sang phải
độ lớn: 2000N ứng vs 2 đoạn, mỗi đoạn 1000N

            3000N ứng vs 3 đoạn, mỗi đoạn 1000N
hình 3 
biểu đạt tại A
phương ngang
chiều: từ trái sang phải
độ lớn: 15N ứng vs 3 đoạn, mỗi đoạn 5N

b: Xét tứ giác ABCK có 

M là trung điểm của AC

M là trung điểm của BK

Do đó: ABCK là hình bình hành

Suy ra: BC=AK