Question

Ex 3 la algebra, dau ​

Answer 1

Răspuns:

Explicație pas cu pas:

Metoda rationalizarii numitorilor:

3√5/2√3 = 3√5·√3/(2·√3·√3) = 3√15/6 = √15/2

2√6/3√2 = 2√6·√2/(3·√2·√2) = 2·√12/6 = 4√3/6 =2√3/3

-7√15/√20 = -7√15/2√5 = -7√15·√5/(2·√5·√5) = -35·√3/10 = -7√3/2

8√15/√20 = 8√15/2√5 = 8√15·√5/(2·5) = 8·5·√3/(2·5) = 4√3

√35/√125 = √35/5√5 = √35·√5/(5·√5·√5) = 5√7/25 = √7/5

√21/√63 = √21/3√7 = √21·√7/(3·√7·√7) = 7√3/21 = √3/3

√6/2√2 = √6·√2/(2·√2·√2) = 2√3/4 = √3/2

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Aceste exercitii se pot rezolva mai usor asa :

2√6/3√2 = 2√(6:2)/3 = 2√3/3

-7√15/√20 = -7√(15:5)/√(20:5) = -7√3/2

8√15/√20 = 8√(15:5)/√(20:5) = 8√3/2 = 4√3

√35/√125 = √7·√5/5√5 = √7/5

√21/√63 = 1/√3 = √3/3

√6/2√2 = √(6:2)/2 = √3/2