Question

Calculați: radical din 10 ori 4 radical din 6; radical din 11 - 6 radical din 2; radical din 7 + 4 radical din 3; radical din 9 + 4 radical din 2

Answer 1

Explicație pas cu pas:

este "minus" la primul radical

$\sqrt{10 - 4 \sqrt{6} } = \sqrt{6 - 2 \cdot \sqrt{6} + 4} = \sqrt{ {( \sqrt{6} )}^{2} - 2 \cdot \sqrt{6} + {2}^{2} } = \sqrt{ {( \sqrt{6} - 2)}^{2} } = |\sqrt{6} - 2| = \bf \sqrt{6} - 2$

$\sqrt{11 - 6 \sqrt{2} } = \sqrt{9 - 2 \cdot 3\sqrt{2} + 2} = \sqrt{ {3}^{2} - 2 \cdot 3\sqrt{2} + {( \sqrt{2} )}^{2} } = \sqrt{ {(3 - \sqrt{2})}^{2} } = |3 - \sqrt{2}| = \bf 3 - \sqrt{2}$

$\sqrt{7 + 4 \sqrt{3} } = \sqrt{4 + 2 \cdot 2\sqrt{3} + 3} = \sqrt{ {2}^{2} + 2 \cdot 2\sqrt{3} + {( \sqrt{3} )}^{2} } = \sqrt{ {(2 + \sqrt{3})}^{2} } = |2 + \sqrt{3}| = \bf 2 + \sqrt{3}$

$\sqrt{9 + 4 \sqrt{2} } = \sqrt{8 + 2 \cdot 2\sqrt{2} + 1} = \sqrt{ {(2 \sqrt{2} )}^{2} + 2 \cdot 2\sqrt{2} + {1}^{2} } = \sqrt{ {(2\sqrt{2} + 1)}^{2} } = |2\sqrt{2} + 1| = \bf 2\sqrt{2} + 1$

Answer 2

$\bf \sqrt{a\pm\sqrt b}=\sqrt{\dfrac{a+c}{2}}\pm\sqrt{\dfrac{a-c}{2}},\ \ unde\ \ c=\sqrt{a^2-b},\ \ a,\ b,\ c\in\mathbb{N}$

$\it 1)\ \ \sqrt{10-4\sqrt6}=\sqrt{10-\sqrt{96}}\\ \\ \sqrt{10^2-96}=\sqrt{100-96}=\sqrt4=2\\ \\ \\ \sqrt{10-4\sqrt6}=\sqrt{\dfrac{10+2}{2}}-\sqrt{\dfrac{10-2}{2}}=\sqrt6-2$

$\it 2)\ \ \sqrt{11-6\sqrt2}=\sqrt{11-\sqrt{72}}\\ \\ \sqrt{11^2-72}=\sqrt{121-72}=\sqrt{49}=7\\ \\ \\ \sqrt{11-6\sqrt2}=\sqrt{\dfrac{11+7}{2}}-\sqrt{\dfrac{11-7}{2}}=3-\sqrt2$