Question

Помогите пожалуйста! Срочно!!! Даю 20 баллов

Answer 1

Ответ:

1.

а)184

б)29

2.

а)7+2√10

б)-10+2√21

Answer 2

Ответ:

1) 1. 184

  2. 29

2) 1. 21+6$\sqrt{10}$

  2. -10+2$\sqrt{21}$

Объяснение:

1) 1. =$3^{2}$- 2*3*5*$\sqrt{7}$+$(5\sqrt{7}) ^{2}$ + 30

  2. =$3^{2}$ + 2*3*2*$\sqrt{5}$+$(2\sqrt{5}) ^{2}$ - 12

2)  1. $\frac{\sqrt{45} +\sqrt{18} }{\sqrt{5} -\sqrt{2} } = \frac{\sqrt{9*5} +\sqrt{9*2} }{\sqrt{5} -\sqrt{2} } = \frac{3*(\sqrt{5}+\sqrt{2} )}{\sqrt{5}-\sqrt{2}} = \frac{3*(\sqrt{5}+\sqrt{2} )*(\sqrt{5}+\sqrt{2} )}{(\sqrt{5}-\sqrt{2})*(\sqrt{5}+\sqrt{2})} = \frac{3*(\sqrt{5}+\sqrt{2} )^{2} }{(\sqrt{5}) ^{2}-(\sqrt{2})^{2} } = \frac{3*(5+2\sqrt{5}\sqrt{2}+2)}{5-4} = \frac{3*(7+2\sqrt{10}) }{1} = 21+6\sqrt{10}$2. $\frac{\sqrt{48} -\sqrt{112} }{\sqrt{3} +\sqrt{7} } = \frac{\sqrt{16*3} -\sqrt{16*7} }{\sqrt{3} +\sqrt{7} } = \frac{4*(\sqrt{3}-\sqrt{7} )}{\sqrt{3}+\sqrt{7}} = \frac{4*(\sqrt{3}-\sqrt{7} )*(\sqrt{3}-\sqrt{7} )}{(\sqrt{3}+\sqrt{7})*(\sqrt{3}-\sqrt{7})} = \frac{4*(\sqrt{3}-\sqrt{7} )^{2} }{(\sqrt{3}) ^{2}-(\sqrt{7})^{2} } = \frac{4*(3-2\sqrt{3}\sqrt{7}+7)}{3-7} = \frac{4*(10-2\sqrt{21}) }{-4} = -10+2\sqrt{21}$