Question

Допоможіть будь ласка

Answer 1

$\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25} \times \sqrt{ 3} = 5 \sqrt{3}$

$2 \sqrt{3} = \sqrt{ {2}^{2} \times 3} = \sqrt{4 \times 3} = \sqrt{12}$

$3 \sqrt{2} = \sqrt{ {3}^{2} \times 2} = \sqrt{9 \times 2} = \sqrt{18}$

$2 \sqrt{3} = \sqrt{ {2}^{2} \times 3} = \sqrt{4 \times 3} = \sqrt{12}$

$\sqrt{18} больше \sqrt{12} , значит3 \sqrt{2} больше2 \sqrt{3}$

4. а)

$\sqrt{100} + \sqrt{49} - \sqrt{16} = \sqrt{ {10}^{2} } + \sqrt{ {7}^{2} } - \sqrt{ {4}^{2} } = 10 + 7 - 4 = 13$

б)

$\sqrt{98} - \sqrt{50} = \sqrt{49 \times 2} - \sqrt{25 \times 2} = \sqrt{ {7}^{2} \times 2} - \sqrt{ {5}^{2} \times 2 } = 7 \sqrt{2} - 5 \sqrt{2} = 2 \sqrt{2}$

в)

$\sqrt{2} \times \sqrt{18} + \frac{ \sqrt{27} }{ \sqrt{3} } = \sqrt{2 \times 18} + \sqrt{ \frac{27}{3} } = \sqrt{36} + \sqrt{9} = \sqrt{9 \times 4} + \sqrt{9} = \sqrt{9} \times \sqrt{4} + \sqrt{9} = 3 \times 2 + 3 = 6 + 3 = 9$

г)

$( \sqrt{8} - \sqrt{32} ) \times \sqrt{2} = \sqrt{8 \times 2} - \sqrt{32 \times 2} = \sqrt{16} - \sqrt{64} = 4 - 8 = - 4$

д)

${( \sqrt{5} - 2 )}^{2} + 4 \sqrt{5} = {( \sqrt{5} )}^{2} - 4 \sqrt{5} + {2}^{2} + 4 \sqrt{5} = 5 - 4 \sqrt{5} + 4 + 4 \sqrt{5} = 5 + 4 = 9$

5. а)

$\frac{24}{ \sqrt{2} } = \frac{ \sqrt{ {24}^{2} } }{ \sqrt{2} } = \sqrt{ \frac{24 \times 24}{2} } = \sqrt{ \frac{12 \times 2 \times 24}{2} } = \sqrt{12 \times 24} = \sqrt{12 \times 12 \times 2} = \sqrt{ {12}^{2} \times 2 } = \sqrt{ {12}^{2} } \times \sqrt{2} = 12 \sqrt{2}$

б)