My name is Khanh , i'm 8 year old . today i'm good.my hobby is cooking.yes i do

-My name is Hoang
-I'm 11 years old.
-I'm happy
- My hobby is playing football
- Yes, I do

1 I'm 8 years old

2 My hobby is cooking

3 yes , i do

4 i play with my friend

- I'm 11 years old.
- My hobby is playing football.
-Yes, I do.
-I play with my friends.
 

1 No, I don't

2 I do crossword

3 It is blue

4 It is playing badminton

1. No,I don't

2. I do word puzzle at break time

3. It's blue

4. I like badmintoning

Tốc độ khi lên dốc:

${\text{v}}^1=\genfrac{}{}{0ex}{}{{\text{v}}^2}{2}=\genfrac{}{}{0ex}{}{60}{2}=30$ km/h

Tốc độ khi xuống dốc:

${\text{v}}^3=1,5{\text{v}}^2=1,5.60=90$ km/h

Độ dài cung đường trên là: 

$s=s^1+s^2+s^3={\text{v}}^1{t}^1+{\text{v}}^2{t}^2+{\text{v}}^3{t}^3=30.\genfrac{}{}{0ex}{}{1}{2}+60.\genfrac{}{}{0ex}{}{1}{6}+90.\genfrac{}{}{0ex}{}{1}{6}=40$ km

$t^1$ = 30 min = 1/2  h ; ${\text{v}}^1=\genfrac{}{}{0ex}{}{{\text{v}}^2}{2}$
${\text{v}}^2$ = 60 km/h ; $t^2$ = 10 min = $\genfrac{}{}{0ex}{}{1}{6}$ h 
$t^3$ = 10 min = $\genfrac{}{}{0ex}{}{1}{6}$ h ; ${\text{v}}^3=1,5{\text{v}}^2$
$s$ = ? (km)

Giải:

Tốc độ khi lên dốc:

${\text{v}}^1=\genfrac{}{}{0ex}{}{{\text{v}}^2}{2}=\genfrac{}{}{0ex}{}{60}{2}=30$ km/h

Tốc độ khi xuống dốc:

${\text{v}}^3=1,5{\text{v}}^2=1,5.60=90$ km/h

Độ dài cung đường trên là: 

$s=s^1+s^2+s^3={\text{v}}^1{t}^1+{\text{v}}^2{t}^2+{\text{v}}^3{t}^3=30.\genfrac{}{}{0ex}{}{1}{2}+60.\genfrac{}{}{0ex}{}{1}{6}+90.\genfrac{}{}{0ex}{}{1}{6}=40$ km

1.Nice to meet you.
2.It's a flower.
3. This is a dog.
4. Touch your head.
5. I can see a bird.

1. Nice to meet you.
2. It's a flower.
3. This is a dog.
4. Touch your head.
5. I can see a bird.

1. Boy

2. Book

3. Arm

4. Red

5. Cat

6. Bear

7. Girl

8. E

Boy

Book

Arm

Red

Cat

Bear

Sister

E

1.It's blue.
2. This is a door.
3. This is my food.
4. I like birds.
5. I can hop.

It's blue

This is a door

This is my food

I like birds

I can hop

a) x - 3 = (3 - x)²

x - 3 = (x - 3)²

x - 3 - (x - 3)² = 0

(x - 3)[1 - (x - 3)] = 0

(x - 3)(1 - x + 3) = 0

(x - 3)(4 - x) = 0

x - 3 = 0 hoặc 4 - x = 0

*) x - 3 = 0

x = 3

*) 4 - x = 0

x = 4

Vậy x = 3; x = 4

b) x³ + 3/2 x² + 3/4 x + 1/8 = 1/64

(x + 1/2)³ = 1/64

(x + 1/2)³ = (1/4)³

x + 1/2 = 1/4

x = 1/4 - 1/2

x = -1/2

a) Ta có: $x-3={\left(3-x\right)}^2$

$\left(x-3\right)-{\left(x-3\right)}^2=0$

$\left(x-3\right)\left(4-x\right)=0$

$x\in \left\{3;4\right\}$.

b) Ta có: $x^3+\genfrac{}{}{0ex}{}{3}{2}{x}^2+\genfrac{}{}{0ex}{}{3}{4}x+\genfrac{}{}{0ex}{}{1}{8}=\genfrac{}{}{0ex}{}{1}{64}$

${\left(x+\genfrac{}{}{0ex}{}{1}{2}\right)}^3={\left(\genfrac{}{}{0ex}{}{1}{4}\right)}^3$

$x+\genfrac{}{}{0ex}{}{1}{2}=\genfrac{}{}{0ex}{}{1}{4}$

$x=\genfrac{}{}{0ex}{}{-1}{4}$.

a) x² + 2xy + y² - x - y

= (x² + 2xy + y²) - (x + y)

= (x + y)² - (x + y)

= (x + y)(x + y + 1)

b) 2x³ + 6x² + 12x + 8

= 2(x³ + 3x² + 6x + 4)

= 2(x³ + x² + 2x² + 2x + 4x + 4)

= 2[(x³ + x²) + (2x² + 2x) + (4x + 4)]

= 2[x²(x + 1) + 2x(x + 1) + 4(x + 1)]

= 2(x + 1)(x² + 2x + 4)