Question

Решите, пожалуйста
Проверить справедливость равенств

Answer 1

1)$(4+ sqrt{15})( sqrt{10}- sqrt{6}) sqrt{4- sqrt{15} }= \ \ = (4+ sqrt{15}) sqrt{( sqrt{10}- sqrt{6}) ^{2}( 4- sqrt{15}) } = \ \ = (4+ sqrt{15}) sqrt{(10-2 sqrt{60}+6) ( 4- sqrt{15}) } = \ \ = (4+ sqrt{15}) sqrt{(16-4 sqrt{15}) ( 4- sqrt{15}) } = \ \ =(4+ sqrt{15}) sqrt{4(4- sqrt{15}) ( 4- sqrt{15}) } = \ \ = (4+ sqrt{15}) 2cdot sqrt{(4- sqrt{15}) ^{2} } =2(4+ sqrt{15})(4-sqrt{15})=2$
2)$frac{1}{ sqrt{7}- sqrt{6} }= frac{sqrt{7}+sqrt{6}}{ (sqrt{7}- sqrt{6})(sqrt{7}+ sqrt{6}) }=frac{sqrt{7}+sqrt{6}}{ (sqrt{7})^2-( sqrt{6})^2 }=frac{sqrt{7}+sqrt{6}}{ 7-6 }=frac{sqrt{7}+sqrt{6}}{ 1}= sqrt{7}+ sqrt{6}$
$frac{3}{ sqrt{6}- sqrt{3} } + frac{4}{ sqrt{7} + sqrt{3} } = frac{3(sqrt{6}+sqrt{3}) }{ (sqrt{6}- sqrt{3})(sqrt{6}+sqrt{3})} + frac{4(sqrt{7} - sqrt{3})}{ (sqrt{7} + sqrt{3})(sqrt{7} - sqrt{3}) } = \ \ =frac{3(sqrt{6}+sqrt{3}) }{ (sqrt{6}) ^{2} - (sqrt{3})^{2} } + frac{4(sqrt{7} - sqrt{3})}{ (sqrt{7}) ^{2} -( sqrt{3}) ^{2} } = frac{3(sqrt{6}+sqrt{3}) }{ 3 } + frac{4(sqrt{7} - sqrt{3})}{4 } = \ \ = sqrt{6}+sqrt{3}+ sqrt{7}- sqrt{3} = sqrt{6}+ sqrt{7}$
левая часть равна правой, равенство доказано.
3)$( frac{ sqrt{2} -1}{ sqrt{2}+1 })^{3} = frac{( sqrt{2}) ^{3} -3( sqrt{2}) ^{2}+3 sqrt{2}-1}{(sqrt{2}) ^{3} +3( sqrt{2}) ^{2}+3 sqrt{2}+1} = frac{ 2sqrt{2} -3cdot 2+3 sqrt{2}-1}{2sqrt{2} +3 cdot 2+3 sqrt{2}+1} = frac{5 sqrt{2}-7 }{5 sqrt{2}-7 }= frac{10-7 sqrt{2} }{10+7 sqrt{2} }$
Значит
$sqrt[3]{ frac{10-7 sqrt{2} }{10+7 sqrt{2} } }= frac{ sqrt{2}-1 }{ sqrt{2}+1 }$