\(5+2\sqrt{6}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)

\(6+2\sqrt{5}=\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(5+2\sqrt{6}=\left(\sqrt{3}+\sqrt{2}\right)^2\)

\(6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)

1) \(2x+5\sqrt{x}-7=2\left[\left(x+\dfrac{5}{2}\sqrt{x}+\dfrac{25}{16}\right)-\dfrac{25}{16}-\dfrac{7}{2}\right]\)

\(=2\left[\left(\sqrt{x}+\dfrac{5}{4}\right)^2-\dfrac{81}{16}\right]=2\left(\sqrt{x}+\dfrac{5}{4}-\dfrac{9}{4}\right)\left(\sqrt{x}+\dfrac{5}{4}+\dfrac{9}{4}\right)=2\left(\sqrt{x}-1\right)\left(\sqrt{x}+\dfrac{7}{2}\right)\)

2) \(3x-7\sqrt{x}+4=3\left[\left(x-\dfrac{7}{3}\sqrt{x}+\dfrac{49}{36}\right)-\dfrac{49}{36}+\dfrac{4}{3}\right]\)

\(=3\left[\left(\sqrt{x}-\dfrac{7}{6}\right)^2-\dfrac{1}{36}\right]=3\left(\sqrt{x}-\dfrac{7}{6}-\dfrac{1}{6}\right)\left(\sqrt{x}-\dfrac{7}{6}+\dfrac{1}{6}\right)=3\left(\sqrt{x}-\dfrac{4}{3}\right)\left(\sqrt{x}-1\right)\)

3) \(4x-4\sqrt{x}-8=4\left[\left(x-\sqrt{x}+\dfrac{1}{4}\right)-\dfrac{1}{4}-2\right]\)

\(=4\left[\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\right]=4\left(\sqrt{x}-\dfrac{1}{2}-\dfrac{3}{2}\right)\left(\sqrt{x}-\dfrac{1}{2}+\dfrac{3}{2}\right)=4\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\)

\(2x+5\sqrt{x}-7=\left(2\sqrt{x}+7\right)\left(\sqrt{x}-1\right)\)

\(3x-7\sqrt{x}+4=\left(\sqrt{x}-1\right)\left(3\sqrt{x}-4\right)\)

\(4x-4\sqrt{x}-8=4\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\)

`11+4sqrt6=8+2.2sqrt2.sqrt3+3=(2sqrt2+sqrt3)^2`

`11-4sqrt6=8-2.2sqrt2.sqrt3+3=(2sqrt2-sqrt3)^2`

`13+4sqrt10=8+2.2sqrt2.sqrt5+5=(2sqrt2+sqrt5)^2`

`13-4sqrt10=8-2.2sqrt2.sqrt5+5=(2sqrt2-sqrt5)^2`

\(11+4\sqrt{6}=\left(\sqrt{8}+\sqrt{3}\right)^2\\ 11-4\sqrt{6}=\left(\sqrt{8}-\sqrt{3}\right)^2\\ 13+4\sqrt{10}=\left(\sqrt{8}+\sqrt{5}\right)^2\\ 13-4\sqrt{10}=\left(\sqrt{8}-\sqrt{5}\right)^2\)

`# \text {04th5}`

`a)`

\(7 \dfrac{3}{5} \div x = 5 \dfrac{4}{15} - 1 \dfrac{1}{6}\)

\(\Rightarrow 7 \dfrac{3}{5} \div x = \dfrac{41}{10}\)

\(\Rightarrow x = 7\dfrac{3}{5} \div \dfrac{41}{10}\)

\(\Rightarrow x = \dfrac{76}{41}\)

Vậy, $x = \dfrac{76}{41}$

`b)`

$x \times 2 \dfrac{2}{3} = 3 \dfrac{4}{8} + 6 \dfrac{5}{12}$

$\Rightarrow x \times \dfrac{2}{3} = \dfrac{119}{12}$

$\Rightarrow x = \dfrac{119}{12} \div \dfrac{2}{3}$

$\Rightarrow x = \dfrac{119}{8}$

Vậy, $x = \dfrac{119}{8}.$

\(P=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}=\dfrac{7-4\sqrt{3}+\sqrt{7-4\sqrt{3}}}{3\sqrt{7-4\sqrt{3}}-1}=\dfrac{7-4\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}}{3\sqrt{\left(2-\sqrt{3}\right)^2}-1}=\dfrac{7-4\sqrt{3}+\left|2-\sqrt{3}\right|}{3\left|2-\sqrt{3}\right|-1}=\dfrac{7-4\sqrt{3}+2-\sqrt{3}}{3\left(2-\sqrt{3}\right)-1}=\dfrac{9-5\sqrt{3}}{5-3\sqrt{3}}=\dfrac{\left(9-5\sqrt{3}\right)\left(5+3\sqrt{3}\right)}{\left(5-3\sqrt{3}\right)\left(5+3\sqrt{3}\right)}=\dfrac{45+2\sqrt{3}-45}{-2}=-\sqrt{3}\)

Thay \(x=7-4\sqrt{3}\) vào P, ta được:

\(P=\dfrac{7-4\sqrt{3}+2-\sqrt{3}}{6-3\sqrt{3}-1}\)

\(=\dfrac{9-5\sqrt{3}}{5-3\sqrt{3}}=-\sqrt{3}\)

\(A=3x^2-12x+16=3\left(x^2-4x\right)+16\)

\(=3\left(x^2-4x+4-4\right)+16\)

\(=3\left(x^2-4x+4\right)-3.4+16\)

\(=3\left(x-2\right)^2+4\ge4\), với mọi x

Vì \(\left(x-2\right)^2\ge0\) với mọi x

nên \(A=3\left(x-2\right)^2+4\ge3.0+4=4\) với mọi x

dấu "=" xảy ra khi và chỉ khi: \(\left(x-2\right)^2=0\Leftrightarrow x=2\)

Vậy giá tri nhỏ nhất của A là 4 tại x=2

\(F=\dfrac{9-x}{x+\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\left(đk:x\ne4\right)=\dfrac{9-x+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{x+\sqrt{x}-6}=\dfrac{9-x+x-9-\left(\sqrt{x}-2\right)^2}{x+\sqrt{x}-6}=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{2-\sqrt{x}}{\sqrt{x}+3}\)

\(F=\dfrac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{9-x+x-9-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\) \(=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{1}{3-\sqrt{x}}\)

Ta có: \(F=\dfrac{9-x}{x+\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\)

\(=\dfrac{9-x+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

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

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

1. was going

2. was attending

3. was standing

4. was riding

5. were living

6. was walking

7. was going

8. was studying / was having

9. were sitting

10. was calling / wasn't / was studying