bài gì vậy nếu được mik làm cho

đc bạn gửi link cho mk đi mk làm cho

Ta có \(|x|=-\frac{5}{3}\)

\(|x|\)≥ 0  mà \(-\frac{5}{3}\)< 0

=> ko tồn tại x ∈ Z t/m với đề bài

                                                Bài giải

Vì : \(\left|x\right|\ge0>-\frac{5}{3}\)

\(\Rightarrow\text{ }x\in\varnothing\)

mình có thấy bài đâu

bài đâu z??

Part 6 : 

1. especially

2. with

3. goods 

4. perform

Part 7: 

1. T

2. F

3. F

4.T

\(1,\\ a,A_1=\left(x-2\right)^2+5\ge5\)

Dấu \("="\Leftrightarrow x=2\)

\(A_2=\left(x+1\right)^2+7\ge7\)

Dấu \("="\Leftrightarrow x=-1\)

\(A_3=\left(3-2x\right)^2-1\ge-1\)

Dấu \("="\Leftrightarrow x=\dfrac{3}{2}\)

\(A_4=\left(x-2\right)^2-3\ge-3\)

Dấu \("="\Leftrightarrow x=2\)

\(b,B_1=\left|x-2\right|+3\ge3\)

Dấu \("="\Leftrightarrow x=2\)

\(B_2=\left|x+1\right|+3\ge3\)

Dấu \("="\Leftrightarrow x=-1\)

\(B_3=\left|2x-4\right|-3\ge-3\)

Dấu \("="\Leftrightarrow x=2\)

\(B_4=\left|6x+1\right|-20\ge-20\)

Dấu \("="\Leftrightarrow x=-\dfrac{1}{6}\)

Bài 1: 

a: \(A_1=\left(x-2\right)^2+5\ge5\forall x\)

Dấu '=' xảy ra khi x=2

\(A_2=\left(x+1\right)^2+7\ge7\forall x\)

Dấu '=' xảy ra khi x=-1

\(A_3=\left(3-2x\right)^2-1\ge-1\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

\(A_4=\left(x-2\right)^2-3\ge-3\forall x\)

Dấu '=' xảy ra khi x=2

1.needn't

2.impressed

3.the

4.send

5.the

6.who

ko biết đúng ko nữa 

Part 1

1 needn't

2 impressed

3 with

4 send

5 the

6 who

Part 2

1f 2d 3c 4b 5a 6g

Part 3

1 was cooking

2 surfing

3 went

4 was going to visit

5 try

6 widened

7 cultural

8 conveniently

Part 4

1 serious -> seriously

2 well-preserved -> be well-preserved

Part 5

1 my english were good

2 going to the english-

3 area has been spoiled 

4 which I read 

Part 6 

1 without

2 be damaged

3 healthy

4 destruction

Part 7

2F 3F 4F

chin;chant;cheek;cheat;chop

flash;dash;fish;bush;rush

tenth;that;this;thin;twelfth

1 B

2 A

3 C

4 B

5 B

6 B

7 B

8 C

9 D

10 B