Автор - diadlo

номера обведённые кружочком,плиз!!

Ответ

Автор - hlopushinairina
3).   (3√2/3-√24+√6)(2√2/3+3√3/2)=
=3·2·2/3+3·3√2/3·√3/2-√24·2√2/3-√24·3√3/2+√6·2√2/3+√6·3√3/2=
=4+9-2√16-3√36+2·√4+3·√9=4+9-8-18+4+9=0;
2)((ab·√a²·√b²/√b²)·b/a²):1/√b²):b/a=
=((ab·a·b·b/(b·a²))·b·a/b=a²b³/ba² ·a=ab²;
3)√(6+2√(5-√(13+√48)))=√(6+2√(5-√(1+4√3+12)))=√(6+2√(5-√(1+2√3)²)))=
=√(6+2√(5-(1+2√3))=√(6+2√(4-2√3))=√(6+2√(3-2√3+1))=√(6+2√(√3-1)²)=
=√(6+2·(√3-1))=√(6+2√3-2)=√(4+2√3)=√(3+2√3+1)=√(√3+1)²=√3+1;

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