Question

Решите все внятно и понятно, 15б!

Answer 1

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

Answer 2

B1.

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

B2.

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