Question

СРОЧНО УМОЛЯЮ ПОЖАЛУЙСТА

Answer 1

Ответ:

1.

$0.4 \sqrt{10} \times \sqrt{250} + \sqrt{169} = 33 \\ 0.4 \times 10 \times 5 + 13 = 20 + 13 = 33$

$\sqrt{24} - \sqrt[4]{6} + \sqrt{54} = ( \sqrt[4]{54} - \sqrt[4]{24} ) \times ( \sqrt[4]{54} + \sqrt[4]{24} ) = \sqrt{6}$

2.

$\sqrt{88} > 9 \\ 2 \sqrt{11} > 3 \sqrt{5}$

4.

$(2 \: \sqrt[]{50} - 5 \sqrt{2} ) \times \sqrt{2} = 10 \\ (10 \sqrt{2} - 5 \sqrt{2} ) \sqrt{2} = 5 \sqrt{2} \sqrt{2} \\ 5 \times 2 = 10$

$(4 \sqrt{75} + 2 \sqrt{12} ) \div 2 \sqrt{3} = 24 \sqrt{3} \\ (20 \sqrt{3} + 4 \sqrt{3} ) \div 2 \times 2 \\ 24 \sqrt{3} \div 2 \times 2 = 12 \sqrt{3} \times 2 = 24 \sqrt{3}$

$\frac{4 - c}{c + 2 \sqrt{c} } = \frac{2 \sqrt{c} - c }{c} = \sqrt{2}$

$\frac{a - 3 \sqrt{a} }{a - 9} = \frac{ \sqrt{a} }{ \sqrt{a + 3} } = \sqrt{3}$

$\frac{ 2\sqrt{11} - 2 }{ \sqrt{33} - \sqrt{3} } = \frac{0}{30}$