Question

Restrângeți după ce ați scos factorii de sub radical:

a)
$\sqrt{45} + \sqrt{20} - \sqrt{125 =} =$
b)
$\sqrt{8} + \sqrt{18} - \sqrt{32}$
c)
$2 \sqrt{160} - 3 \sqrt{90 } + 5 \sqrt{40}$
d)
$\sqrt{27} + \sqrt{243} + \sqrt{75} - \sqrt{108}$
e)
$8 \sqrt{216} - 3 \sqrt{96} + \sqrt{54} - 2 \sqrt{150} + \sqrt{600}$
f)
$0.2 \sqrt{75} - \frac{5}{4} \sqrt{48} + 2.(6) \times \sqrt{27}$
​

Answer 1

$\sqrt{45} +\sqrt{20} -\sqrt{125} =\\3\sqrt{5} +2\sqrt{5} -5\sqrt{5} =\\5\sqrt{5} -5\sqrt{5} =0\\\\\sqrt{8} +\sqrt{18} -\sqrt{32} =\\2\sqrt{2} +3\sqrt{2} -4\sqrt{2} =\\5\sqrt{2} -4\sqrt{2} =\sqrt{2} \\\\2\sqrt{160} -3\sqrt{90} +5\sqrt{40} =\\2*4\sqrt{10} -3*3\sqrt{10} +5*2\sqrt{10} =\\8\sqrt{10} -9\sqrt{10} +10\sqrt{10} =\\-\sqrt{10} +10\sqrt{10} =9\sqrt{10}$

$\sqrt{27} +\sqrt{243} +\sqrt{75} -\sqrt{108} =\\3\sqrt{3} +9\sqrt{3} +5\sqrt{3} -6\sqrt{3} =\\12\sqrt{3} +5\sqrt{3} -6\sqrt{3} =\\17\sqrt{3} -6\sqrt{3} =11\sqrt{3} \\\\8\sqrt{216} -3\sqrt{96} +\sqrt{54} -2\sqrt{150} +\sqrt{600} =\\8*6\sqrt{6} -3*4\sqrt{6} +3\sqrt{6} -2*5\sqrt{6} +10\sqrt{6} =\\48\sqrt{6} -12\sqrt{6} +3\sqrt{6} -10\sqrt{6} +10\sqrt{6} =\\48\sqrt{6} -12\sqrt{6} +3\sqrt{6}=\\36\sqrt{6} +3\sqrt{6} =39\sqrt{6}$

$0.2\sqrt{75} -\frac{5}{4} \sqrt{48} +2.(6)*\sqrt{27} \\\sqrt{(0,2)^2*75} -1,25\sqrt{48} +\frac{8}{3}\sqrt{27} =\\\sqrt{3} -\sqrt{75} +\frac{8}{3}*3\sqrt{3} =\\\sqrt{3} -5\sqrt{3} +8\sqrt{3} =\\-4\sqrt{3} +8\sqrt{3}=4\sqrt{3}$

Answer 2

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