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`

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

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

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

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

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

`12+2sqrt35=7+2sqrt{7.5}+5=(sqrt7+sqrt5)^2`

`9+4sqrt2=8+2.2sqrt2+1=(2sqrt2+1)^2`

`9-4sqrt2=8-2.2sqrt2+1=(2sqrt2-1)^2`

\(12+2\sqrt{35}=\left(\sqrt{7}+\sqrt{5}\right)^2\)

\(9+4\sqrt{2}=\left(2\sqrt{2}+1\right)^2\)

\(9-4\left(\sqrt{2}\right)=\left(2\sqrt{2}-1\right)^2\)

`# \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, $x = \dfrac{119}{8}.$

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

Dễ ẹc = 3,2  và 0 chứ dề

= 3,2

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