How long has she researched on lung cancer?

careful 

Thay \(x=-2\) vào phương trình, ta có

\(a\left(-2\right)^2-2b\left(-2\right)+3=0\)

\(\Leftrightarrow4\left(a+b\right)=-3\)

\(\Leftrightarrow a+b=-\dfrac{3}{4}\) (1)

Thay \(x=1\) vào phương trình, ta có

\(a.1^2-2b.1+3=0\)

\(\Leftrightarrow a-2b=-3\) (2)

Trừ (2) cho (1) theo vế, ta được

\(-3b=-\dfrac{9}{4}\Rightarrow b=\dfrac{3}{4}\) \(\Rightarrow a=-\dfrac{3}{4}-\dfrac{3}{4}=-\dfrac{3}{2}\)

Vậy \(a=-\dfrac{3}{2}\) và \(b=\dfrac{3}{4}\)

2. C → prepare

3. C → lost

4. B → ∅

5. D → had done

6. C → did i find

7. C → couldn't sleep

8. A → is

9. B → had been

Câu 10 mình nhìn k rõ  

Don't worry. She will be there in two hours.

Thanks for your flowers. They really help to cheer me up.

\(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)

\(\Rightarrow7\left(x-1\right)=6\left(x+5\right)\)

\(\Rightarrow7x-7=6x+30\)

\(\Rightarrow x=37\)

Ở đây dùng can't để diễn tả một sự việc chắc chắn không xảy ra, mà tính huống là quá khứ nên dùng can't have been

9. The teacher remarked (that) that student spoke English very well

10. Father said (that) he thought it wouldn't rain the following day

C 

a. Ta có

\(VP=\left(a+b\right)^3-3ab\left(a+b\right)\)

\(=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)

\(=a^3+b^3\) ( đpcm )

b. Ta có

\(VP=\left(a-b\right)^3+3ab\left(a-b\right)\)

\(=a^3-3a^2b+3ab^2-b^3+3a^2b-3ab^2\)

\(=a^3-b^3\) ( đpcm )