Question

Calculați: a 2√3+3.4√3; C √6 (3√5-2√5)-√√2-√15; √3 b 5√2-2√2-2;? d (4√2-√2). √2+√√2. √18; (2√5+3√5)-√5 (√3+2√3); 12(√12-√3)+(√27-√√48)-3








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Answer 1

Răspuns:

Salut

Explicație pas cu pas:

$a) 2\sqrt{3} + 3* 4\sqrt{3} = 2\sqrt{3} + 12\sqrt{3} = 14\sqrt{3}$

$b) 5\sqrt{2} - 2\sqrt{2} * 2 = 5\sqrt{2} - 4\sqrt{2} = \sqrt{2}$

$c) \sqrt{6} * ( 3\sqrt{5} - 2\sqrt{5} ) - \sqrt{2} * \sqrt{15} = \sqrt{6} * \sqrt{5} - \sqrt{30} = \sqrt{30} - \sqrt{30} = 0$

$d) ( 4\sqrt{2} - \sqrt{2} ) * \sqrt{2} + \sqrt{2} * \sqrt{18} = 3\sqrt{2} * \sqrt{2} + \sqrt{36} = 3 * 2 + \sqrt{36} = 6 + \sqrt{36} = 6 + 6 = 12$

$e) \sqrt{3} * ( 2\sqrt{5} + 3\sqrt{5} ) - \sqrt{5} * ( \sqrt{3} + 2\sqrt{3} ) = \sqrt{3} * 5\sqrt{5} - \sqrt{5} * 3\sqrt{3} = 5\sqrt{15} - 3\sqrt{15} = 2\sqrt{15}$

$f) 2 * ( \sqrt{12} - \sqrt{3} ) + ( \sqrt{27} - \sqrt{48} ) * 3 = 2\sqrt{12} - 2\sqrt{3} + 3\sqrt{27} - 3\sqrt{48} = 2 * 2\sqrt{3} - 2\sqrt{3} + 3 * 3\sqrt{3} - 3 * 4\sqrt{3} = 4\sqrt{3} - 2\sqrt{3} + 9\sqrt{3} - 12\sqrt{3} = 2\sqrt{3} + 9\sqrt{3} - 12\sqrt{3} = 11\sqrt{3} - 12\sqrt{3} = -\sqrt{3}$

Sper că ți-am fost de folos!

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