Đặt\(\begin{cases} x+y=S \\ xy=P \end{cases}\)

Ta có:\(\begin{cases} S-2P=0 \\ S-P^2=\sqrt{(P-1)^2+1} \end{cases}\)

\(\Leftrightarrow\)\(\begin{cases} S=2P \\ 2P-P^2=\sqrt{(P-1)^2+1} \end{cases}\)

\(\Leftrightarrow\)\(\begin{cases} S=2P \\ (2P-P^2)^2=(P-1)^2+1 \end{cases}\)

\(\Leftrightarrow\)\(\begin{cases} S=2P \\ 4P^2-4P^3+P^4=P^2-2P+2 \end{cases}\)

\(\Leftrightarrow\)\(\begin{cases} S=2P \\ P^4-4P^3+3P^2+2P-2=0 \end{cases}\)

\(\Leftrightarrow\)\(\begin{cases} S =2+2\sqrt{3}\\ P=1+\sqrt{3} \end{cases}\)(1)hoặc\(\begin{cases} S=2 \\ P=1 \end{cases}\)(2)hoặc\(\begin{cases} S=2-2 \sqrt{3}\\ P=1-\sqrt{3} \end{cases}\)(3)

Còn lại là thay vào biểu thức x²-Sx+P=0 thôi

However

effect on

to help

such

in

from

I

1 will have

2 would have

3 have

4 had

5 will make

6 would make

7 has

8 had

9 is

10 were

II

1 have

2 had

3 had had

4 will go

5 would go

6 would have gone

7 is

8 were - would visit

9 had been - would have visited

10 wouldn't be - were

I.

1. will have

2. would have

3. have

4. had

5. will made

6. would made

7. has

8. had

9. is

10. were

II.

1. have

2. had

3. had had

4. will go

5. would go

6. would have gone

7. is

8. were - would visit

9. had been - would have visited

10. will not be - is

Bài 2 :

a, Ta có đồ thị :

b, Ta có : \(\tan a=3\)

\(\Rightarrow a\approx71,5^o\)

KO BIẾT LÀM NHA BẠN!!!!!!!!!!!!!!!!

Bài 3:

(1) \(N_2+3H_2⇌2NH_3\)

(2) \(NH_3+HNO_3\rightarrow NH_4NO_3\)

(3) \(NH_4NO_3+KOH\rightarrow KNO_3+NH_3+H_2O\)

(4) \(N_2+O_2\xrightarrow[]{t^ocao}2NO\)

(5) \(NO+\dfrac{1}{2}O_2\rightarrow NO_2\)

(6) \(4NO_2+O_2+2H_2O\rightarrow4HNO_3\)

(7) \(4HNO_{3\left(đ\right)}+Cu\rightarrow Cu\left(NO_3\right)_2+2NO_2+2H_2O\)

(8) \(HNO_3+NH_3\rightarrow NH_4NO_3\)

Bài 4.

Ta có số mol khí:
$ n_{NO_2}=\dfrac{4{,}48}{22{,}4}=0{,}2\ \text{mol} $

Phương trình hóa học:
$ Cu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O $

$ Al + 6HNO_3 \rightarrow Al(NO_3)_3 + 3NO_2 + 3H_2O $

Gọi:
$ x $ là số mol $Cu$
$ y $ là số mol $Al$

Theo phương trình phản ứng:
$ 2x + 3y = 0{,}2 \quad (1) $

Theo khối lượng hỗn hợp:
$ 64x + 27y = 3 \quad (2) $

Giải hệ (1) và (2) thu được:
$ x = 0{,}05 \ \text{mol} $
$ y = 0{,}0333 \ \text{mol} $

Khối lượng mỗi kim loại:
$ m_{Cu} = 0{,}05 \times 64 = 3{,}2,\text{g} $

$ m_{Al} = 0{,}0333 \times 27 = 0{,}9,\text{g} $

Phần trăm khối lượng:

%$Cu = \dfrac{3{,}2}{3} \times 100% \approx 66{,}7 $%

%$Al = \dfrac{0{,}9}{3} \times 100% = 30 $%

THAM KHẢO:

1 B

2 C

3 C

4 D

5 A

6 A

7 C

8 C

9 B

10 D

11 D

12 B

13 B

14 D

15 A

16 B

17 D

18 D

19 C

20 D

21 C

22 B

23 B

24 B

25 A

26 D

27 A