Помогите! Разложите на множители: а) √12+√15. б) 7-4у^2 в) m+2√mn+n г) 2√b+10√a-√ab-5a. Created by masha751137505. matematika-ru

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

Rechnung

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

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Artem112

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

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