1. These sentences have been made up to illustrate how different phrasal verbs based on "do" and "make" are used.
2. Why did Bob make off so quickly when I arrived?
3. We're doing up our kitchen, and we could do with more paint.
4. Buying a house is expensive. We'll have to do without a holiday this year.
5. A new house should make for no holiday.
6. We're doing away with our very old cooker and buying a new one.
7. John sent me a note yesterday, and I can't make out what he's written.
8. We didn't have a map, so we made for the hills hoping to find somewhere to stay.

Ta có a: 6 dư 5
=> a= 6k+5 với k ϵ N
có: a² = (6k+5)² = 36k²+ 60k+25
vì 36k²⋮6 ; 60k⋮6 ; 25 : 6 dư 1
=> a² chia 6 dư 1 
 

Lời giải:

Vì $a$ chia $6$ dư $5$ nên đặt $a=6k+5$ với $k$ nguyên. 

Khi đó: $a^2=(6k+5)^2=36k^2+25+60k=6(6k^2+10k+4)+1$ chia $6$ dư $1$

\((x-y+2)^2\\=x^2+y^2+2^2-2xy-2\cdot y\cdot2+2\cdot x\cdot2\\=x^2+y^2+4-2xy-4y+4x\)

1 We've them out of run

2 If you across it come tell me

3 you should look up it

4 My friend came up with it

5 so don't throw them away

1 run our of them

2 if you come across it

3 you should look it up

4 my friend came up with it

5 so don't throw them away

1 B

2 D

3 B

4 A

5 B

6 C

7 A

\(\dfrac{4x+2}{4x-2}+\dfrac{3-6x}{6x-6}\left(dkxd:x\ne\dfrac{1}{2};x\ne1\right)\)

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

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

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

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

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

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

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

\(=\dfrac{2x-3}{4x^2-6x+2}\)

\(x\) + 2y = 8

\(2y\)        = 8 - \(x\)

 y        = \(\dfrac{8-x}{2}\)

  y =  - \(\dfrac{x}{2}\) + 4

Thay y = - \(\dfrac{x}{2}\) + 4 vào biểu thức B = \(xy\) ta có: 

B = \(x\).(-\(\dfrac{x}{2}\) + 4)

B = - \(\dfrac{x^2}{2}\) + 4\(x\)

B = -\(\dfrac{1}{2}\). (\(x^2\)  - 8\(x\)  + 16)  +  8 

B = - \(\dfrac{1}{2}\).(\(x\) - 4)² + 8

Vì  \(\dfrac{1}{2}\).(\(x\) - 4)² ≥ 0 ⇒ - \(\dfrac{1}{2}\).(\(x\) - 4)² ≤ 0 ⇒ - \(\dfrac{1}{2}\).(\(x\)  - 4)² + 8 ≤ 8

Dấu bằng xảy ra khi:  \(x\) - 4 = 0 ⇒ \(x\) = 4; thay \(x\) = 4 vào biểu thức:

y = - \(\dfrac{1}{2}\) \(x\)+ 4 ta có y = - \(\dfrac{4}{2}\) + 4 = 2

Vậy giá trị lớn nhất của B là 8 xảy ra khi \(x\) = 4; y = 2