Question

14. Aratati ca urmatoarele numere sunt rationale:


Am nevoie urgenta, dau coroana!​

Answer 1

Răspuns:

6-4√2=4-4√2+2=(2-√2)²

√(2-√2)²=l2-√2l=2-√2

6+4√2=4+4√2+2=(2+√2)²

√(2+√2)²=l2+√2l=2+√2

2-√2+2+√2=4∈Q

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b)14-6√5=9-6√5+5=(3-√5)²

√(3-√5)²=l3-√5l=3-√5

14+6√5=9+6√5+5=(3+√5)²

√(3+√5)²=l3+√5l=3+√5

3-√5+3+√5=6

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c)12-6√3=9-6√3=3=(3-√3)²

√(3-√3)²=l3-√3l=3-√3

12+6√3=....=3+√3

3-√3+3+√3=6

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Explicație pas cu pas:

Answer 2

$\it b)\ \sqrt{14-6\sqrt5}=\sqrt{9+5-6\sqrt5}=\sqrt{3^2-6\sqrt5+(\sqrt5)^2}=\sqrt{(3-\sqrt5)^2}=\\ \\ =|3-\sqrt5|\\ \\ Analog,\ \sqrt{14+6\sqrt5}=|3+\sqrt5|\\ \\ \sqrt{14-6\sqrt5}+\sqrt{14+6\sqrt5}=|3-\sqrt5|+|3+\sqrt5|=3-\sqrt5+3+\sqrt5=6=\\ \\ =\dfrac{6}{1}\in\mathbb{Q}$

$\it f)\ \sqrt{9+4\sqrt5}= \sqrt{5+4+4\sqrt5}=\sqrt{(\sqrt5)^2+4\sqrt5+2^2}=\sqrt{(\sqrt5+2)^2}=\\ \\ =|\sqrt5+2| =\sqrt5+2\\ \\ Analog,\ \sqrt{9-4\sqrt5}=\sqrt5-2\\ \\ \sqrt{9+4\sqrt5}-\sqrt{9-4\sqrt5}=\sqrt5+2-(\sqrt5-2)=\sqrt5+2-\sqrt5+2=4=\\ \\ =\dfrac{4}{1}\in \mathbb{Q}$