Question

Помогите! Разложите на множители: а) √12+√15. б) 7-4у^2 в) m+2√mn+n г) 2√b+10√a-√ab-5a

Answer 1

$a) sqrt{12}+ sqrt{15}= sqrt{4*3}+ sqrt{5*3}= sqrt{4}* sqrt{3}+ sqrt{5}* sqrt{3}=\\= sqrt{3}(2+ sqrt{5}) \\b)7-4y^2=( sqrt{7})^2-(2y)^2=( sqrt{7}-2y)( sqrt{7}+2y)\\c)m+2 sqrt{mn}+n=( sqrt{m})^2+2 sqrt{mn}+( sqrt{n})^2=( sqrt{m}+ sqrt{n})^2\\d)2 sqrt{b}+10 sqrt{a}- sqrt{ab}-5a=(2 sqrt{b}- sqrt{a}* sqrt{b} )+(10 sqrt{a}-5( sqrt{a})^2)=\\= sqrt{b}(2- sqrt{a})+5 sqrt{a}(2- sqrt{a})=(2- sqrt{a})( sqrt{b}+5 sqrt{a})$

Answer 2

$sqrt{12}+ sqrt{15} = sqrt{3}( sqrt{4}+ sqrt{5})=sqrt{3}( 2+ sqrt{5}) \ 7-4y^2=(sqrt{7}-2y)(sqrt{7}+2y) \ m+2 sqrt{mn} +n=(sqrt{m}+sqrt{n})^2 \ 2 sqrt{b} +10 sqrt{a}- sqrt{ab}-5a=sqrt{b}(2-sqrt{a})+5sqrt{a}(2-sqrt{a})= \ =(2-sqrt{a})(sqrt{b}+5sqrt{a})$