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Lyanaaa
@Lyanaaa
November 2021
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а) 3√5+√20+√80
б) избавиться от иррациональности в знаменателе дроби: 2/3-√2х-1
номер 2
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армения20171
А)3√5+√20+√80=3√5+√(4•5)+√(16•5)=
3√5+√4•√5+√16√5=3√5+2√5+4√5=
9√5
б)2/(3-√(2х-1))=
2(3+√(2х-1))/(3-√(2х-1))(3+√(2х-1))=
2(3+√(2х-1))/(3²-√(2х-1)²)=
2(3+√(2х-1)/(8-2х)=(3+√(2х-1))/(4-х)
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Answers & Comments
3√5+√4•√5+√16√5=3√5+2√5+4√5=
9√5
б)2/(3-√(2х-1))=
2(3+√(2х-1))/(3-√(2х-1))(3+√(2х-1))=
2(3+√(2х-1))/(3²-√(2х-1)²)=
2(3+√(2х-1)/(8-2х)=(3+√(2х-1))/(4-х)