Question

ПОМОГИТЕ СРОЧНО ПОЖАЛУЙСТА ​

Answer 1

$\displaystyle\bf\\1)\\\\\frac{a-25}{a-5\sqrt{a} } =\frac{(\sqrt{a})^{2} -5^{2} }{(\sqrt{a} )^{2}-5\sqrt{a} } =\frac{(\sqrt{a}-5)\cdot(\sqrt{a}+5) }{\sqrt{a} \cdot(\sqrt{a} -5)} =\frac{\sqrt{a} +5}{\sqrt{a} } \\\\\\2)\\\\\frac{x-4\sqrt{x} \sqrt{y} +4y}{x-4y} =\frac{(\sqrt{x} )^{2} -2\cdot \sqrt{x} \cdot 2\sqrt{y} +(2\sqrt{y} )^{2} }{(\sqrt{x} )^{2} -(2\sqrt{y} )^{2} } =$

$\displaystyle\bf\\=\frac{(\sqrt{x} -2\sqrt{y} )^{2} }{(\sqrt{x} -2\sqrt{y} )\cdot(\sqrt{x} +2\sqrt{y} )}=\frac{\sqrt{x} -2\sqrt{y} }{\sqrt{x} +2\sqrt{y} } \\\\\\3)\\\\\frac{11+\sqrt{22} }{\sqrt{22} +2} =\frac{(\sqrt{11} )^{2} +\sqrt{11\cdot 2} }{\sqrt{11\cdot 2} +(\sqrt{2} )^{2} } =\frac{\sqrt{11} \cdot(\sqrt{11} +\sqrt{2}) }{\sqrt{2} \cdot(\sqrt{11} +\sqrt{2} } =\\\\\\=\frac{\sqrt{11} }{\sqrt{2} } =\sqrt{5,5}$

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