1 have to

2 must

3 mustn't

4 have to

1 doesn't have to       2 have to
3 have to                   4 don't have to
5 mustn't                   6 have to

1 doesn't have to

2 must

3 have to

4 don't have to

5 mustn't

6 must

1. My sister can swim well. She can't play badminton. (but)

à………My sister can swim well, but she can't play badminton.………………………………………………………………………………

2. Hanh goes to that French restaurant every week. She loves French food. (so)

à……Hanh loves French food, so she goes to that French restaurant every week……………………………………………………………………………………

3. He was tired last night. He stayed up late to watch a football match. (although)

à……Although he was tired last night, he stayed up late to watch a football match…………………………………………………………………………………

4. I don't like this TV programme. It's so boring. (because)

à……I don't like this TV programme because it's so boring……………………………………

5. My mum went to Hue last month. His mum went to Hue last month, too.(and)

à………My mum and his mum went to Hue last month…………………………………………………………………………………

6. He was lazy, so he got a bad mark.  ( “because”)

à……Because he was lazy, he got a bad mark…………………………………………………………………………………

7. It rained hard yesterday. I went to school on time. (“although”)

à……Although it rained hard yesterday, I went to school on time…………………………………………………………………………………

8. Daisy  has to do her homework. She is very tired .            ( “Although”)

……Although Daisy is very tired, she has to do her homework………………………………………………………………………………………

tác giả của tác phẩm chí phèo là nhà văn Nam Cao đó

a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-21}{7}=-3\)

=>\(x=-3\cdot2=-6\)

\(A=5x+4x=9x=9\cdot\left(-6\right)=-54\)

b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{y-x}{5-3}=\dfrac{4}{2}=2\)

=>\(\left\{{}\begin{matrix}x=2\cdot3=6\\y=2\cdot5=10\end{matrix}\right.\)

\(B=3x+5y=3\cdot6+5\cdot10=18+50=68\)

Ta có:

\(A=1+1-1+1-1+1-1\ldots\)

Bây giờ, ta tính \(1-A\) :

\(1-A=1-\left(1-1+1-1+1-1\ldots\right)\)

Bỏ ngoặc ta có:

\(1-\left(1-1+1-1\ldots\right)=1-1+1-1+1\ldots\)

Ta nhận thấy vế bên phải chính là \(A\) nên ta có:

\(1-A=A\)

A⋮B

=>\(x^4-3x^3+x^2+a\cdot x+b-1\) ⋮\(x^2+x+1\)

=>\(x^4+x^3+x^2-4x^3-4x^2-4x+4x^2+4x+4+a\cdot x+b-5\) ⋮\(x^2+x+1\)

=>a=0 và b-5=0

=>a=0 và b=5