3. Peter apologised for having broken the glass.

4. You shouldn't work too hard.

5. The security guard told us to keep away from the sea when we approached the fence.

6. She asked him: "Have you studied French?".

7. He told me to bring swimming things in case it was sunny.

8. The lady - reseacher advised us to take science appreciation courses at school.

9. The man asked me how many students were here in my class.

10. The doctor advised Mr. Rogerts to take more exercise.

8. Jack gives Thien An 5 million dong a month.

9. Does Messy play football very well?

10. Does carrot contain a lot of vitamins?

c) Điều kiện xác định: \(\left\{{}\begin{matrix}x\ge0\\x+3\sqrt{x}-10\ne\\\sqrt{x}+5\ne0\end{matrix}\right.0\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

 \(C_3=\dfrac{5\sqrt{x}-3}{x+3\sqrt{x}-10}-\dfrac{4}{\sqrt{x}+5}\)

\(=\dfrac{5\sqrt{x}-3}{\left(x+5\sqrt{x}\right)-\left(2\sqrt{x}+10\right)}-\dfrac{4}{\sqrt{x}+5}\)

\(=\dfrac{5\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}+5\right)-2\left(\sqrt{x}+5\right)}-\dfrac{4}{\sqrt{x}+5}\)

\(=\dfrac{5\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+5\right)}-\dfrac{4\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+5\right)}\)

\(=\dfrac{5\sqrt{x}-3-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+5\right)}\)

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

\(=\dfrac{1}{\sqrt{x}-2}\)

D. not old enough

\(C=\dfrac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}+\sqrt{7-4\sqrt{3}}+\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)

\(=\left(\dfrac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}+\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}\right)+\left(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\right)\)

\(=\left[\dfrac{\left(\sqrt{7}-\sqrt{3}\right)^2}{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}+\dfrac{\left(\sqrt{7}+\sqrt{3}\right)^2}{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}\right]+\left(\sqrt{4-4\sqrt{3}+3}+\sqrt{4+4\sqrt{3}+3}\right)\)

\(=\dfrac{\left(\sqrt{7}-\sqrt{3}\right)^2+\left(\sqrt{7}+\sqrt{3}\right)^2}{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}+\left[\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\right]\)

\(=\dfrac{7-2\sqrt{7}.\sqrt{3}+3+7+2\sqrt{7}.\sqrt{3}+3}{7-3}+\left[\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\right]\)

\(=\dfrac{20}{4}+\left(\left|2-\sqrt{3}\right|+\left|2+\sqrt{3}\right|\right)\)

\(=5+2-\sqrt{3}+2+\sqrt{3}\) (vì \(2-\sqrt{3}>0,\)  \(2+\sqrt{3}>0\))

\(=9\)

1. \(A=\sqrt{8}+\sqrt{32}-6\sqrt{2}\)

\(=\sqrt{4.2}+\sqrt{16.2}-6\sqrt{2}\)

\(=\sqrt{4}.\sqrt{2}+\sqrt{16}.\sqrt{2}-6\sqrt{2}\)

\(=2\sqrt{2}+4\sqrt{2}-6\sqrt{2}\)

\(=\left(2+4-6\right)\sqrt{2}\)

\(=0\)

a) 

\(x+3+\sqrt{x^2-6x+9}\)

\(=x+3+\sqrt{x^2-2.3.x+3^2}\)

\(=x+3+\sqrt{\left(x-3\right)^2}\)

\(=x+3+\left|x-3\right|\)

\(=x+3+3-x\) (vì \(x\le3\) nên \(x-3\le0\))

\(=6\)

b)

\(\sqrt{x^2+4x+4}-\sqrt{x^2}\)

\(=\sqrt{x^2+2.2.x+2^2}-\left|x\right|\)

\(=\sqrt{\left(x+2\right)^2}+x\) (vì \(x\le0\))

\(=\left|x+2\right|+x\)

\(=x+x+2\) (vì \(x\ge-2\) nên \(x+2\ge0\))

\(=2x+2\)

4.
a) Điều kiện xác định: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}+2\ne0\\\sqrt{x}-2\ne0\\x-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

\(A=\dfrac{4}{\sqrt{x}+2}+\dfrac{2}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-6}{x-4}\)

\(=\dfrac{4\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{5\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{4\left(\sqrt{x}-2\right)+2\left(\sqrt{x}+2\right)-\left(5\sqrt{x}-6\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{4\sqrt{x}-8+2\sqrt{x}+4-5\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{1}{\sqrt{x}-2}\)

b) Thay \(x=\dfrac{1}{9}\) (thỏa mãn điều kiện xác định) vào biểu thức  \(A,\)  ta có:

\(A=\dfrac{1}{\sqrt{\dfrac{1}{9}}-2}=\dfrac{1}{\dfrac{1}{3}-2}=\dfrac{1}{\dfrac{1-6}{3}}=\dfrac{1}{\dfrac{-5}{3}}=-\dfrac{3}{5}\)

Điều kiện: \(\left\{{}\begin{matrix}3x-2\ge0\\7-x\ge0\end{matrix}\right.\Leftrightarrow\dfrac{2}{3}\le x\le7\)
\(\sqrt{3x-2}-\sqrt{7-x}+3x^2-20x=-9\)

\(\Leftrightarrow\left(\sqrt{3x-2}-4\right)-\left(\sqrt{7-x}-1\right)+\left(3x^2-20x+12\right)=0\)

\(\Leftrightarrow\dfrac{\left(3x-2\right)-16}{\sqrt{3x-2}+4}-\dfrac{\left(7-x\right)-1}{\sqrt{7-x}+1}+\left[\left(3x^2-18x\right)-\left(2x-12\right)\right]=0\)

\(\Leftrightarrow\dfrac{3x-18}{\sqrt{3x-2}+4}+\dfrac{x-6}{\sqrt{7-x}+1}+\left[3x\left(x-6\right)-2\left(x-6\right)\right]=0\)

\(\Leftrightarrow\dfrac{3\left(x-6\right)}{\sqrt{3x-2}+4}+\dfrac{x-6}{\sqrt{7-x}+1}+\left(3x-2\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left[\dfrac{3}{\sqrt{3x-2}+4}+\dfrac{1}{\sqrt{7-x}+1}+\left(3x-2\right)\right]=0\)

\(\Leftrightarrow x-6=0\) (do \(\dfrac{3}{\sqrt{3x-2}+4}+\dfrac{1}{\sqrt{7-x}+1}+\left(3x-2\right)>0\))

\(\Leftrightarrow x=6\) (thỏa mãn)

Vậy phương trình có tập nghiệm là:  \(S=\left\{6\right\}.\)

Bài 7:
\(M=\left(\dfrac{1}{x-4}-\dfrac{1}{x+4\sqrt{x}+4}\right).\dfrac{x+2\sqrt{x}}{\sqrt{x}}\)

\(=\left[\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\)

\(=\left[\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}\right]\left(\sqrt{x}+2\right)\)

\(=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\left(\sqrt{x}+2\right)\)

\(=\dfrac{4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}\)

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