Question

Номер 427(б,г,е);428;429(а,в,д)...................

Answer 1

$alpha ^2- beta ^2=( alpha - beta )( alpha + beta ) \ sqrt{ alpha beta } = sqrt{ alpha } cdot sqrt{ beta }$

$5-c^2=( sqrt{5} -c)( sqrt{5} +c) \ 11-16b^2=( sqrt{11} -4b)( sqrt{11} +4b) \ x-y=( sqrt{x} - sqrt{y} )( sqrt{x} + sqrt{y} )$

$3+ sqrt{3} = sqrt{3} ( sqrt{3} +1) \ 10-2 sqrt{10} = sqrt{10} ( sqrt{10} -2) \ sqrt{x} +x= sqrt{x} (1+ sqrt{x} ) \ a-5 sqrt{a} = sqrt{a} ( sqrt{a} -5) \ sqrt{a} - sqrt{2a} = sqrt{a} (1- sqrt{2} ) \ sqrt{3m} + sqrt{5m} = sqrt{m} ( sqrt{3} + sqrt{5} ) \ sqrt{14} - sqrt{7} =sqrt{7} (sqrt{2} -1) \ sqrt{33} +sqrt{22} =sqrt{11} (sqrt{3}+sqrt{2} )$

$frac{b^2-5}{b- sqrt{5} } = frac{(b- sqrt{5})(b+ sqrt{5})}{b- sqrt{5} } =b+ sqrt{5} \ frac{2- sqrt{x} }{x-4} =- frac{ sqrt{x} -2}{(sqrt{x} -2)(sqrt{x} +2)} =- frac{1}{sqrt{x} +2} \ frac{a-b}{ sqrt{b}+ sqrt{a} } = frac{( sqrt{a}- sqrt{b})( sqrt{a}+ sqrt{b})}{ sqrt{b}+ sqrt{a} } = sqrt{a}- sqrt{b}$