Điều kiện xác định: \(\left\{{}\begin{matrix}25-x^2\ge0\\15-x^2\ge0\end{matrix}\right.\Leftrightarrow-\sqrt{15}\le x\le\sqrt{15}\)

\(\sqrt{25-x^2}-\sqrt{15-x^2}=-2\)

\(\Leftrightarrow\sqrt{25-x^2}+2=\sqrt{15-x^2}\)

\(\Leftrightarrow\left(\sqrt{25-x^2}+2\right)^2=\left(\sqrt{15-x^2}\right)^2\)

\(\Leftrightarrow25-x^2+4\sqrt{25-x^2}+4=15-x^2\)

\(\Leftrightarrow14+4\sqrt{25-x^2}=0\) 

Vì \(14+4\sqrt{25-x^2}\ge14\) nên phương trình đã cho vô nghiệm.

Theo giả thiết, ta có:

- \(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)

\(\Rightarrow x^3=\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)^3\)

\(=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{9+4\sqrt{5}}.\sqrt[3]{9-4\sqrt{5}}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)

\(=18+3\sqrt[3]{81-80}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)

\(=18+3.\sqrt[3]{1}.x\)

\(=18+3x\)

- \(y=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)

\(\Rightarrow y^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\)

\(=3+2\sqrt{2}+3-2\sqrt{2}+3\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\)

\(=6+3\sqrt[3]{9-8}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\)

\(=6+3\sqrt[3]{1}.y\)

\(=6+3y\)

Do đó: \(P=x^3+y^3-3\left(x+y\right)+1993\)

\(=18+3x+6+3y-3x-3y+1993\)

\(=2017\)

The test was being done at 3pm yesterday.

56 từ 

1. was => were

2. had => have

3. worked => work

4. you didn't => didn't you

5. hated => hate

6. be => bỏ

7. readed => read

8. did go => went

a) \(A=\sqrt{4x+20}-2\sqrt{x+5}+\sqrt{9x+45}\)

\(=\sqrt{4\left(x+5\right)}-2\sqrt{x+5}+\sqrt{9\left(x+5\right)}\)

\(=2\sqrt{x+5}-2\sqrt{x+5}+3\sqrt{x+5}\)

\(=3\sqrt{x+5}\)

b) Để A = 6 thì \(3\sqrt{x+5}=6\Leftrightarrow\sqrt{x+5}=2\Leftrightarrow x+5=4\Leftrightarrow x=-1\) (nhận)

Vậy với x = -1 thì A = 6.

Điều kiện xác định: \(\left\{{}\begin{matrix}x-2000\ge0\\y-2001\ge0\\z-2002\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2000\\y\ge2001\\z\ge2002\end{matrix}\right.\)

\(\sqrt{x-2000}+\sqrt{y-2001}+\sqrt{z-2002}=\dfrac{1}{2}\left(x+y+z\right)-3000\)

\(\Leftrightarrow2\sqrt{x-2000}+2\sqrt{y-2001}+2\sqrt{z-2002}=\left(x+y+z\right)-6000\)

\(\Leftrightarrow\left(x-2000-2\sqrt{x-2000}+1\right)+\left(y-2001-2\sqrt{y-2001}+1\right)+\left(z-2002-2\sqrt{z-2002}+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x-2000}-1\right)^2=0\\\left(\sqrt{y-2001}-1\right)^2=0\\\left(\sqrt{z-2002}-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2000}-1=0\\\sqrt{y-2001}-1=0\\\sqrt{z-2002}-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2000}=1\\\sqrt{y-2001}=1\\\sqrt{z-2002}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2000=1\\y-2001=1\\z-2002=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2001\\y=2002\\z=2003\end{matrix}\right.\) (nhận)

Vậy \(\left(x;y;z\right)=\left(2001;2002;2003\right)\) là nghiệm của phương trình.

13. beauty

14. introduce

15. famous

16. different

17. industry

18. compulsory

31. Singapore is one of the countries of Association of South East Asian Nations.

32. What were you doing when the mailman came to your house?

19. B

20. A

21. C

22. D

23. A

24. D

1. B

2. C

3. A

4. A

5. D

6. A

7. B

8. C

9. C

10. D