Question

Решите по братски @@@@@@@@@@@@@@@@@

Answer 1

Відповідь:

9.

$1) \frac{\sqrt{\sqrt{3} -1} *\sqrt{\sqrt{3} +1} }{3\sqrt{2} } = \frac{\sqrt{ \sqrt{3}^{2} -1^{2} } }{3\sqrt{2} } = \frac{\sqrt{3-1} }{{3\sqrt{2} } } = \frac{\sqrt{2} }{3\sqrt{2} } =\frac{1}{3} \\\\2) \frac{\sqrt{\sqrt{5} -\sqrt{2} } *\sqrt{\sqrt{5} +\sqrt{2} } }{2\sqrt{3} } = \frac{\sqrt{(\sqrt{5}) ^{2} -(\sqrt{2} ^{2} )} }{2\sqrt{3} } = \frac{\sqrt{5-2} }{2\sqrt{3} } = \frac{\sqrt{3} }{2\sqrt{3} } = \frac{1}{2}$

$3) \frac{3}{\sqrt{\sqrt{7} -2} *\sqrt{\sqrt{7} -2}}=\frac{3}{\sqrt{(\sqrt{7} )^{2} - 2^{2} } } = \frac{3}{\sqrt{7-4} } =\frac{3}{\sqrt{3} } \\\\4) \frac{2}{\sqrt{\sqrt{7} -\sqrt{5} } *\sqrt{\sqrt{7} +\sqrt{5} } } =\frac{2}{\sqrt{(\sqrt{7} )^{2}-(\sqrt{5} )^{2} } } = \frac{2}{\sqrt{7-5} } = \frac{2}{\sqrt{2} }$

10.

$1) \frac{4b^{2} }{a^{4} } = \frac{2b}{a^{4} } \\\\\\\\2)\frac{\sqrt{9x^{6} } }{y^{2} } = \frac{3x^{3} }{y^{2} } \\\\\\\3) \frac{\sqrt{16} }{x^{4} y^{2} } = \frac{4}{x^{4} y^{2}} \\\\\\4) \frac{\sqrt{25} }{a^{2}b^{6} } = \frac{5}{a^{2}b^{6}}$

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