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

Bài 4: 

\(x=130^0\)

`@` `\text {Ans}`

`\downarrow`

`4,`

Vì `\text {MN // BC}`

`=>` $\widehat {B} = \widehat {BMN} = 114^0 (\text {2 góc đối đỉnh})$

Ta có: \(\left\{{}\begin{matrix}\widehat{\text{BMN}}+\widehat{\text{AMN}}=180^0\left(\text{2 góc kề bù}\right)\\\widehat{\text{CNM}}+\widehat{\text{ANM}}=180^0\left(\text{2 góc kề bù}\right)\end{matrix}\right.\)

`=>`\(\left\{{}\begin{matrix}\widehat{\text{AMN}}=180^0-114^0=66^0\\\widehat{\text{ANM}}=180^0-130^0=50^0\end{matrix}\right.\)

Xét `\Delta AMN`:

\(\widehat{\text{A}}+\widehat{\text{M}}+\widehat{\text{N}}=180^0\left(\text{định lý tổng 3 góc trong 1 tgiac}\right)\)

`=>`\(\widehat{\text{A}}+66^0+50^0=180^0\)

`=>`\(\widehat{\text{A}}=180^0-66^0-50^0=64^0\)

Mà \(\widehat{\text{A}}=\widehat{\text{x}}\)

`=>`\(\widehat{\text{x}}=64^0\)

Vậy, số đo của góc `x = 64^0.`

1 were - would you play

2 weren't studying - would have 

3 had taken - wouldn't have got

4 would you go - could

5 will you give - is

6 recycle - won't be

7 had heard - wouldn't have gone

8 would you buy - had

9 don't hurry - will miss

10 had phoned - would have given

11 were - wouldn't eat

12 will go - rains

13 had known - would have sent

14 won't feel - swims

15 hadn't freezed - would have gone

\(a,=\dfrac{x^3+2x}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}=\dfrac{x^3+2x+2x-2-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^3+3x-3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^3+3}{\left(x^2+x+1\right)}\)

cho mik câu trả lời chii tiết nhất với ạ !!

b,=\(\dfrac{x+1}{x+3}+\dfrac{x-1}{x-3}+\dfrac{2x-2x^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x-3\right)+\left(x-1\right)\left(x+3\right)+2x-2x^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+x-3x-3+x^2-x+3x-3+2x-2x^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+2}{x+3}\)

em chịu_{em thua}

cái gì đây?

của bạn nè.Mik lớp 5 nhưng vẫn phải học thuộc hết 

2 A

3 D

4 A

5 A

6 B

7 A

8 A

9 A

a: \(=\dfrac{x^3+2x+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{x^3-x^2+3x-3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+3}{x^2+x+1}\)

b: \(=\dfrac{x^2-2x-3+x^2+2x-3+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)

c: \(=\dfrac{6-7+x}{3\left(x-1\right)}=\dfrac{x-1}{3\left(x-1\right)}=\dfrac{1}{3}\)

d: \(=\dfrac{x^3+2x+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^3-x^2+3x-3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+3}{x^2+x+1}\)

1 is explained

2 was stolen

3 will be opened

4 is being closed

5 is going to be built

1C

2A

3C

4B

5C

6D

7B

8A

9D

10C