Question

21. Calculați a + b şi 2a + 3b, unde: √12 a=(√32-√√72+√128-√200)
şi
b = √8 (√48-√27+√192-√108). ​.

Answer 1

Explicație pas cu pas:

$a = \sqrt{12} \cdot (\sqrt{32} - \sqrt{72} + \sqrt{128} - \sqrt{200}) =$

$= 2\sqrt{3} \cdot (4 \sqrt{2} - 6 \sqrt{2} + 8 \sqrt{2} - 10 \sqrt{2})$

$= 2 \sqrt{3} \cdot (- 4 \sqrt{2}) = \bf - 8 \sqrt{6}$

$b = \sqrt{8} \cdot ( \sqrt{48} - \sqrt{27} + \sqrt{192} - \sqrt{108}) =$

$= 2\sqrt{2} \cdot ( 4\sqrt{3} - 3\sqrt{3} + 8\sqrt{3} - 6\sqrt{3})$

$= 2\sqrt{2} \cdot 3\sqrt{3} = \bf 6 \sqrt{6}$

=>

$a + b = - 8 \sqrt{6} + 6 \sqrt{6} = \bf - 2 \sqrt{6}$

$2a + 3b = 2 \cdot (- 8 \sqrt{6}) + 3 \cdot 6 \sqrt{6} =\\= - 16 \sqrt{6} + 18 \sqrt{6} = \bf 2 \sqrt{6}$