\(2 \sqrt{\frac{2}{3}} - 4 \sqrt{\frac{3}{2}}\)

\(a)=2\frac{\sqrt2}{\sqrt{3}}-4\frac{\sqrt{3}}{\sqrt{2}}\)

\(=\frac{2\sqrt2}{\sqrt{3}}-\frac{4\sqrt{3}}{\sqrt{2}}\)

\(=\left(\frac{2\sqrt2}{\sqrt{3}}\right)\cdot\sqrt3-\left(\frac{4\sqrt{3}}{\sqrt{2}}\right)\cdot\sqrt2\)

\(=\frac{2\sqrt6}{3}-2\sqrt6\)

\(=\frac{2\sqrt6-2\sqrt6\cdot3}{3}\)

\(=\frac{\left(2-2\cdot3\right)\sqrt6}{3}\)

\(=\frac{-4\sqrt6}{3}\)

\(b)\frac{5 \sqrt{48} - 3 \sqrt{27} + 2 \sqrt{12}}{\sqrt{3}}\)

\(=\frac{20\sqrt3-9\sqrt3+4\sqrt3}{\sqrt{3}}\)

\(=\frac{\left(20-9+4\right)\sqrt3}{\sqrt{3}}\)

\(=\frac{15\sqrt3}{\sqrt{3}}\)

\(=15\)

\(c)\frac{1}{3 + 2 \sqrt{2}}+\frac{4 \sqrt{2} - 4}{2 - \sqrt{2}}\)

\(=\frac{3-2\sqrt{2}}{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}+\frac{\left(4\sqrt{2}-4\right)\left(2+\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}\)

\(=\frac{3-2\sqrt{2}}{3^2-\left(2\sqrt2\right)^2}+\frac{4\sqrt2}{2^2-\left(\sqrt2\right)^2}\)

\(=\frac{3-2\sqrt{2}}{9-8}+\frac{4\sqrt2}{4-2}\)

\(=3-2\sqrt{2}+2\sqrt2\)

\(=3\)

\(a)\left(\right.\sqrt{\frac{4}{3}}+\sqrt{3}\left.\right).\sqrt{6}\)

\(=\left(\right.\frac{2\sqrt3}{3}+\sqrt{3}\left.\right).\sqrt{6}\)

\(=\left(\right.\frac{2\sqrt3}{3}+\frac{3\sqrt3}{3}\left.\right).\sqrt{6}\)

\(=\frac{5\sqrt3}{3}.\sqrt{6}\)

\(=\frac{5\sqrt{18}}{3}\)

\(=\frac{5\cdot3\sqrt2}{3}\)

\(=5\sqrt2\)

\(b)\left(\right.1-2\sqrt{5}\left.\right)^2\)

\(=1^2-2\cdot1\cdot2\sqrt5+\left(2\sqrt5\right)^2\)

\(=1-4\sqrt5+20\)

\(=21-4\sqrt5\)

\(c)2\sqrt{3}-\sqrt{27}\)

\(=2\sqrt{3}-3\sqrt3\)

\(=-\sqrt3\)

\(d)\sqrt{45}-\sqrt{20}+\sqrt{5}\) 

\(=3\sqrt5-2\sqrt5+\sqrt{5}\)

\(=\left(3-2+1\right)\sqrt{5}\)

\(=2\sqrt{5}\)

\(a)A=\frac{\sqrt{3}}{\sqrt{\sqrt{3} + 1} - 1}-\frac{\sqrt{3}}{\sqrt{\sqrt{3} + 1} + 1}\)

\(A=\frac{\sqrt3\left(\sqrt{\sqrt{3} + 1}+1\right)}{\left(\sqrt{\sqrt{3} + 1}-1\right)\left(\sqrt{\sqrt{3} + 1}+1\right)}-\frac{\sqrt{3}\left(\sqrt{\sqrt{3} + 1}-1\right)}{\left(\sqrt{\sqrt{3} + 1}-1\right)\left(\sqrt{\sqrt{3} + 1}+1\right)}\)

\(A=\frac{\sqrt3\left(\sqrt{\sqrt{3} + 1}+1\right)-\sqrt{3}\left(\sqrt{\sqrt{3} + 1}-1\right)}{\sqrt{\left(\sqrt3+1\right)^2}-1^2}\)

\(A=\frac{2\sqrt3}{\sqrt3+1-1^{}}\)

\(A=\frac{2\sqrt3}{\sqrt3^{}}\)

\(A=2\)

\(b)B=\left(\right.\frac{15}{\sqrt{6} + 1}+\frac{4}{\sqrt{6} - 2}-\frac{12}{3 - \sqrt{6}}\left.\right)\left(\right.\sqrt{6}+11\left.\right)\)

\(B=\left\lbrack\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\frac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\frac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right\rbrack\left(\right.\sqrt{6}+11\left.\right)\)

\(B=\left\lbrack\frac{15\left(\sqrt{6}-1\right)}{6-1}+\frac{4\left(\sqrt{6}+2\right)}{6-4}-\frac{12\left(3+\sqrt{6}\right)}{9-6}\right\rbrack\left(\right.\sqrt{6}+11\left.\right)\)

\(B=\left\lbrack3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right\rbrack\left(\right.\sqrt{6}+11\left.\right)\)

\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12+4\sqrt6\right)\left(\sqrt{6}+11\right)\)

\(B=\left\lbrack\sqrt6\left(3+2-4\right)-11\right\rbrack\left(\sqrt{6}+11\right)\)

\(B=\left(\sqrt6-11\right)\left(\sqrt{6}+11\right)\)

\(B=6-121\)

\(B=-115\)

\(c)C=4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\frac{1}{5}}\)

 \(C=4\sqrt{5\cdot4}-3\sqrt{5\cdot25}+5\sqrt{5\cdot9}-15\frac{\sqrt1}{\sqrt5}\)

 \(C=8\sqrt5-15\sqrt5+15\sqrt5-15\frac{\sqrt5}{5}\)

 \(C=8\sqrt5-15\sqrt5+15\sqrt5-3\sqrt5\)

 \(C=8\sqrt5-3\sqrt5\)

 \(C=5\sqrt5\)

I love city life. First, it is very convenient to live in the city. The public transport system reaches almost all areas of the city, so it is easy for me to get around. In addition, there are many shops that sell all kinds of goods, so I can buy almost everything I need. Second, the city often has many good schools and hospitals. Therefore, people here can enjoy quality education and healthcare. Finally, city life is exciting. There are many entertainment places for me and my friends. For example, we can hang out at shopping malls, watch movies at the cinema, and visit beautiful parks downtown. In conclusion, I find the city a liveable place for me.

Gọi:

- Quãng người đó đi xe đạp từ A - B là x (0 < x ; KM)

- Thời gian người đó đi xe đạp từ A - B là \(\dfrac{x}{15}\)(giờ)

- Thời gian người đó đi xe đạp từ B - A là \(\dfrac{x}{12}\)(giờ)

- Đổi 45 phút = \(\dfrac{3}{4}\) giờ

Theo đề bài ra ta có phương trình:

\(\dfrac{x}{12}\)-\(\dfrac{x}{15}\)=\(\dfrac{3}{4}\)

\(\dfrac{5x}{60}\)-\(\dfrac{4x}{60}\)=\(\dfrac{45}{60}\)

5x - 4x = 45

X = 45

Vậy quãng đường người đó đi từ A - B là 45 KM

a) \(\dfrac{ }{ }\)\(\dfrac{3x+15}{x^2-9}\)+\(\dfrac{1}{x+3}\)-\(\dfrac{2}{x-3}\)
= \(\dfrac{3x+15}{\left(x-3\right)\left(x+3\right)}\)+\(\dfrac{x-3}{\left(x-3\right)\left(x+3\right)}\)+\(\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{\text{3 x + 15 + x − 3 − 2 x − 6}}{\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{6+2x}{\left(x-3\right)\left(x+3\right)}\)

=\(\dfrac{2\left(x+3\right)\text{ }}{\left(x-3\right)\left(x+3\right)}\)

=\(\dfrac{2}{x-3}\)