5. C

6. A

7. B

8. C

9. A

10. C

11. C

12. A

1.B

2.C

3.A

4.C

1.B

2.C

3.A

4.C

1. A

2. B

3. D

4. A

5. D

CaCO₃

B

9. A

13. A

16. D

17. D

18. C

13. C

14. D

20. C

\(x^2+\dfrac{1}{x^2}=7\\ \\ \\ \\ \Rightarrow x^2+2x\cdot\dfrac{1}{x}+\dfrac{1}{x^2}=7+2x\cdot\dfrac{1}{x}=9\\ \\ \\ \\ \Rightarrow\left(x+\dfrac{1}{x}\right)^2=9\\ \\ \\ \\ \Rightarrow x+\dfrac{1}{x}=\pm3\)

Trường hợp 1: \(x+\dfrac{1}{x}=3\) ta có:

\(x^3+\dfrac{1}{x^3}=\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)-\left(x+\dfrac{1}{x}\right)=7\cdot3-3=18\)

Trường hợp 2: \(x+\dfrac{1}{x}=-3\) ta có:

\(x^3+\dfrac{1}{x^3}=\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)-\left(x+\dfrac{1}{x}\right)=7\cdot\left(-3\right)-\left(-3\right)=-18\)

1. C

2. B

3. A

4. A

5. B

6. C