giải pt : 2cosx+tan2x=1

\(3\sqrt{x-1}+2\sqrt{4\left(x-1\right)}-3\sqrt{9\left(x-1\right)}+6=0\)

⇔\(3\sqrt{x-1}+4\sqrt{x-1}-9\sqrt{x-1}+6=0\)

⇔\(-2\sqrt{x-1}+6=0\)

1 A: You look hot. What have you been doing (you / do)?

B: I have been running (run).

A: Running! In this heat? How far have you run (you / run)?

B: About four miles.

2 A: What’s the problem? You look a bit preoccupied.

B: Yes, I have been thinking (think) about Helen.

A: Why? Is there something wrong with her?

B: Well, she has been acting ( act ) so strangely lately.

A: In what way?

B: Well, some days when she arrives at work, I know that she has been crying (cry). And she has been making (make) private calls when we’re all out at lunch. I just think that something’s going on.

A: Have you talked (you / talk) to her about it yet?

B: Yes, a few times, and each time she has said (say) that there’s nothing wrong but I’m not so sure.

3 A: Where have you been (you / be), Simon?

B: I have been talking (talk) to Mark on the phone. He says he has been trying (try) to phone us all day.

A: Well I have been (be) in most of the day but I haven't heard (not hear) the phone.

B: That’s strange. But anyway, he has had (have) a phone call from Jackie’s mother and Jackie have been (be) in some kind of accident. It’s nothing very serious but she’s got to stay in hospital overnight.

A: Oh dear. has she broken (she / break) any bones?

B: I’m not quite sure how badly she has been hurt (be hurt) but I think we should go and see her tonight.

(1) has been

(2) hace filled

(3) have always listened

(4) have acted

5/ has interviewing

6/ have you interviewed

7/ have interviewed

8/ have been talking

9/ have been discussing

10/ have had

11/ have also been working

12/ have been

13/ have been circulating

14/ have been seeing

15/ have been learning

16 / haven't learnt

a,CuO=> Cu hóa trị II.

tinh bột (C₆H₁₀O₅)_n

n_{tinh bột}=16,2/162=0,1 (vì trong quá trình làm sẽ triệt tiêu n)

tinh bột-----60%------->2CO₂

0,1------------------------->0,12 (mol)

CO₂----------------------->CaCO₃

0,12----------------------->0,12 (mol)

m_CaCO3=0,12.100=12 (gam)