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15 tháng 11 2023

1 If we continue polluting the air, the environment will be affected seriously

2 If we don't stop damaging birds' homes, there will be fewer birds left

3 If we continue polluting the water, there will be more disease in people and animals

4 If we keep polluting the land, the soil will not be safe enough for us to grow food in

5 If we use public transportation when we need, there will be less air pollution

12 Daniel signs up for a clean-up campaign so that he can join in it

13 People in my neighborhood clean the park up so that their children will have a nice place to play

14 Almost all the students in my school plant trees so that they can have much fresher air

15 tháng 11 2023

Em cảm ơn ạ

12 tháng 11 2023

9. A
10. D
5. A
6. B
7. D
8. B
 

13 tháng 11 2023

9 A

10 D

14 tháng 11 2023

e: \(\dfrac{x^2+3x+9}{x^3+4x^2+4x}\cdot\dfrac{x^2+2x}{x^3-27x}\)

\(=\dfrac{x^2+3x+9}{x\left(x^2+4x+4\right)}\cdot\dfrac{x\left(x+2\right)}{x\left(x^2-27\right)}\)

\(=\dfrac{x^2+3x+9}{\left(x+2\right)^2}\cdot\dfrac{x+2}{x\left(x^2-27\right)}\)

\(=\dfrac{\left(x^2+3x+9\right)}{\left(x+2\right)\cdot x\left(x^2-27\right)}\)

f: \(\dfrac{2x^2+4xy+2y^2}{5x-5y}\cdot\dfrac{15x-15y}{2x^3+2y^3}\)

\(=\dfrac{2\left(x^2+2xy+y^2\right)}{5\left(x-y\right)}\cdot\dfrac{15\left(x-y\right)}{2\left(x^3+y^3\right)}\)

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

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

g: \(\dfrac{x^3-4x}{x^2-7x+12}\cdot\dfrac{x-4}{x^2-2x}\)

\(=\dfrac{x\left(x^2-4\right)}{\left(x-3\right)\left(x-4\right)}\cdot\dfrac{x-4}{x\left(x-2\right)}\)

\(=\dfrac{x^2-4}{\left(x-3\right)\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x-2\right)}=\dfrac{x+2}{x-3}\)

14 tháng 11 2023

Em cảm ơn nhìu ạ 😍❤️

m: \(=x^m\cdot x^2-x^m=x^m\left(x^2-1\right)=x^m\left(x-1\right)\left(x+1\right)\)

n: \(=5\cdot x^m\cdot x^2+10x^2\)

\(=5x^2\left(x^m+2\right)\)

o: \(=5x\left(x-2y\right)+2\left(x-2y\right)^2\)

\(=\left(x-2y\right)\left(5x+2x-4y\right)\)

=(x-2y)(7x-4y)

p: \(=7x\left(y-4\right)^2+\left(y-4\right)^3\)

\(=\left(y-4\right)^2\cdot\left(7x+y-4\right)\)

q: \(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)

\(=\left(4x-8\right)\left(x^2+6-x-7\right)-9\left(4x-8\right)\)

\(=\left(4x-8\right)\left(x^2-x-10\right)\)

\(=4\left(x-2\right)\left(x^2-x-10\right)\)

5 tháng 8 2023

m) \(x^{m+2}-x^m\)

\(=x^m\cdot x^2-x^m\)

\(=x^m\left(x^2-1\right)\)

\(=x^m\left(x^2-1^2\right)\)

\(=x^m\left(x-1\right)\left(x+1\right)\)

n) \(5x^{m+2}+10x^2\)

\(=5x^m\cdot x^2+10x^2\)

\(=5x^2\cdot x^m+10x^2\)

\(=5x^2\left(x^m+2\right)\)

o) \(5x\left(x-2y\right)+2\left(2y-x\right)^2\)

\(=5x\left(x-2y\right)+2\left(x-2y\right)^2\)

\(=\left(x-2y\right)\left[5x+2\left(x-2y\right)\right]\)

\(=\left(x-2y\right)\left(5x+2x-4y\right)\)

\(=\left(x-2y\right)\left(7x-4y\right)\)

p) \(7x\left(y-4\right)^2-\left(4-y\right)^3\)

\(=7x\left(4-y\right)^2-\left(4-y\right)^3\)

\(=\left(4-y\right)^2\left[7x-\left(4-y\right)\right]\)

\(=\left(4-y\right)^2\left(7x-4+y\right)\)

q) \(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)

\(=4\left(x-2\right)\left(x^2+6\right)-4\left(x-2\right)\left(x+7\right)-36\left(x-2\right)\)

\(=4\left(x-2\right)\left[\left(x^2+6\right)-\left(x+7\right)-9\right]\)

\(=4\left(x-2\right)\left(x^2+6-x-7-9\right)\)

\(=4\left(x-2\right)\left(x^2-x-10\right)\)

Bài 7:

a: a>b

nên -2a<-2b

=>-2a-6<-2b-6<-2b

b: a>b

nên 3a>3b

=>3a-9>3b-9

=>3(a-3)>3(b-3)

HQ
Hà Quang Minh
Giáo viên
24 tháng 7 2023

\(x:y:z=\dfrac{4}{7}:\dfrac{5}{8}:\dfrac{9}{10}=160:175:252\)

24 tháng 7 2023

Giúp e vs e đg rất gấp ạ

\(4x^4+1\)

\(=4x^4+4x^2+1-4x^2\)

\(=\left(2x^2+1\right)^2-\left(2x\right)^2\)

\(=\left(2x^2+1+2x\right)\left(2x^2+1-2x\right)\)

NV
4 tháng 1

\(4x^4+1=4x^4+4x^2+1-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2\)

\(=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)

Bài 2: Chọn C

Bài 4: 

a: \(\widehat{C}=180^0-80^0-50^0=50^0\)

Xét ΔABC có \(\widehat{A}=\widehat{C}< \widehat{B}\)

nên BC=AB<AC

b: Xét ΔABC có AB<BC<AC

nên \(\widehat{C}< \widehat{A}< \widehat{B}\)

Cái chỗ này mình xin lỗi bạn nhiều nha, mình bị sai chỗ này rồi

Ta có: \(\left(a-b\right)^2>=0\forall a,b\)

=>\(a^2+b^2-2ab>=0\forall a,b\)

=>\(a^2+b^2>=2ab\forall a,b\)

Dấu "=" xảy ra khi a=b

\(2\cdot\left(\dfrac{a}{2}\cdot\dfrac{b}{2}\right)< =\left(\dfrac{a}{2}\right)^2+\left(\dfrac{b}{2}\right)^2=\dfrac{a^2+b^2}{4}\)

=>\(2\cdot\left(2\cdot\dfrac{a}{2}\cdot\dfrac{b}{2}\right)< =\dfrac{2\left(a^2+b^2\right)}{4}=\dfrac{a^2+b^2}{2}\)

Bạn bỏ giúp mình dấu căn nha

 

1 tháng 11 2023

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