Question

Решите все внятно и понятно, 15б!

Answer 1

$( \sqrt{8 - 2 \sqrt{7} } - \sqrt{8 + 2 \sqrt{7} } )^{2} = {( \sqrt{8 - 2 \sqrt{7} } )}^{2} - 2 \times \sqrt{8 - 2 \sqrt{7} } \times \sqrt{8 + 2 \sqrt{7} } + {( \sqrt{8 + 2 \sqrt{7} } )}^{2} = 8 - 2 \sqrt{7} - 2 \times \sqrt{(8 - 2 \sqrt{7})(8 + 2 \sqrt{7} )} + 8 + 2 \sqrt{7} = 16 - 2 \times \sqrt{ {8}^{2} - {(2 \sqrt{7} )}^{2} } = 16 - 2 \times \sqrt{64 - 28} = 16 - 2 \times \sqrt{36} = 16 - 2 \times 6 = 16 - 12 = 4$
$\frac{1}{2 \sqrt{5} - 4 } - \frac{1}{4 + 2 \sqrt{5} } = \frac{2 \sqrt{5 } + 4 }{(2 \sqrt{5} - 4)( 2\sqrt{5} + 4) } - \frac{2 \sqrt{5} - 4}{(2 \sqrt{5} - 4)(2 \sqrt{5} + 4) } = \frac{2 \sqrt{5} + 4 - 2 \sqrt{5} + 4}{(2 \sqrt{5} - 4)(2 \sqrt{5} + 4)} = \frac{8}{ {(2 \sqrt{5}) }^{2} - {4}^{2} } = \frac{8}{20 - 16} = \frac{8}{4} = 2$

Answer 2

B1.

$(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}})^2 = (\sqrt{7+1-2\sqrt{7}}-\sqrt{7+1+2\sqrt{7}})^2 = (\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1})^2 = (\sqrt{\sqrt{7}^2-2\sqrt{7}*1+1^2}-\sqrt{\sqrt{7}^2+2\sqrt{7}*1+1^2})^2 = (\sqrt{(\sqrt{7}-1)^2} - \sqrt{(\sqrt{7}+1)^2})^2 = (\sqrt{7} -1-(\sqrt{7} +1))^2=(\sqrt{7}-1-\sqrt{7}-1)^2= (-2)^2=4$

B2.

$\frac{1}{2\sqrt{5}-4} -\frac{1}{4+2\sqrt{5}} = \frac{4+2\sqrt{5}-2\sqrt{5}+4}{(2\sqrt{5}-4)(2\sqrt{5}+4)} =\frac{8}{(2\sqrt{5})^2-4^2} =\frac{8}{4*5-16}=\frac{8}{4}=2$