giúp mình bài này với mình cảm ơn
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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
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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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
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Bài 2 :
a, Ta có đồ thị :
b, Ta có : \(\tan a=3\)
\(\Rightarrow a\approx71,5^o\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(HNO_3+NH_3\rightarrow NH_4NO_3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
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