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10radical din1,44+radical din 48-radical din 12-6 supra radical din 3-5 raadical din 0,16

Răspuns :


[tex]\it 10\sqrt{1,44} +\sqrt{48} -\sqrt{12}-\dfrac{^{\sqrt3)}6}{ \ \sqrt3} -5\sqrt{0,16} = 10\sqrt{\dfrac{144}{100}} +\sqrt{16\cdot3}- \\\;\\ \\\;\\ -\sqrt{4\cdot3} - \dfrac{6\sqrt3}{3} - 5\sqrt{\dfrac{ \ 16^{(4}}{100}} =10\sqrt{\dfrac{36}{25}} +4\sqrt3-2\sqrt3-2\sqrt3- \\\;\\ \\\;\\ -5\sqrt{\dfrac{4}{25}} =10\cdot\dfrac{6}{5} -5\cdot\dfrac{2}{5} =12-2=10[/tex]