\(2x\left(x+3\right)-3\left(x^2+1\right)=x+1-x\left(x-2\right)\)

\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)

\(\Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\)

Put the correct form of the verbs in brackets:
1. Nam often (give) gives me stamps from other countries.
2. Does she (go) go to school everyday ?
3. We (live) live near Nga’s house, but we (not see) don't see her very often.
4. Be careful! The teacher (look) is looking at you.
5. The sun ( rise ) rises in the east.
6. The Earth ( go ) goes around the sun. Now, the sun ( shine ) is shining
7. Are you ( write ) writing a letter now?
8. My uncle ( not drink ) doesn't drink beer very often.
9. An ( help ) is helping her father repair the radio at the moment.
10. It is raining (rain) very hard at present.
11.Be quiet ! The baby ( sleep ) is sleeping .
12. The sun ( set) sets in the West.
13. They ( not / have ) aren't having lunch in the restaurant now.
14. Next summer Mike’s parents are taking him to New York to see a new film. (take)
15. What time  does he usually have (have) breakfast?

1. I enjoy reading books very much

2. I used to go to the cinema at weekends

3. Peter is not old enough to see the horror film

4. It takes me three and a half hours to do this exercise

5. It's interesting to see that film

6. The test is too difficult for the pupils to finish in time

Câu 2:
a) Ta có \(x=4-2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-2\right)^2}=\sqrt{3}-2\)

Thay \(x=\sqrt{3}-1\) vào \(B\), ta được

\(B=\dfrac{\sqrt{3}-1-2}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-3}{\sqrt{3}}=1-\sqrt{3}\)

b) Để \(B\) âm thì \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\) mà \(\sqrt{x}+1\ge1>0\forall x\) \(\Rightarrow\sqrt{x}-2< 0\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)

c) Ta có \(B=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=1-\dfrac{3}{\sqrt{x}+1}\)

Với mọi \(x\ge0\) thì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{3}{\sqrt{x}+1}\le3\Rightarrow B=1-\dfrac{3}{\sqrt{x}+1}\ge-2\)

Dấu "=" xảy ra khi \(\sqrt{x}+1=1\Leftrightarrow x=0\)

Vậy \(B_{min}=-2\) khi \(x=0\)

Despite the fact that we had argued, we really get on well with each other again

a. \(\left(x+3\right)^2-9=0\)

\(\Leftrightarrow\left(x+3-3\right)\left(x+3+3\right)=0\)

\(\Leftrightarrow x\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

b. \(x-6=\left(6-x\right)^2\)

\(\Leftrightarrow x-6=\left(x-6\right)^2\)

\(\Leftrightarrow\left(x-6\right)\left(1-x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\7-x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=7\end{matrix}\right.\)

c. \(\left(2x-3\right)^2-\left(3+2x\right)^2=0\)

\(\Leftrightarrow\left(2x-3+3+2x\right)\left(2x-3-3-2x\right)=0\)

\(\Leftrightarrow-24x=0\) \(\Leftrightarrow x=0\)

d. \(x^3-6x^2+12x-8=0\)

\(\Leftrightarrow\left(x-2\right)^3=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)