Question

Спростити вираз
Обчислити його значення, якщо b=16/49

Answer 1

$\displaystyle\bf\\1)\\\\\frac{1}{\sqrt{b} -5} -\frac{\sqrt{b} }{b-25} =\frac{1}{\sqrt{b} -5}-\frac{\sqrt{b} }{(\sqrt{b})^{2} -5^{2} } =\\\\\\=\frac{1}{\sqrt{b} -5} -\frac{\sqrt{b} }{(\sqrt{b}-5)(\sqrt{b}+5) } = \frac{\sqrt{b} +5-\sqrt{b} }{(\sqrt{b} -5)(\sqrt{b}+5) } =\\\\\\=\frac{5}{(\sqrt{b} -5)(\sqrt{b} +5)} \\\\\\2)\\\\\frac{5}{(\sqrt{b} -5)(\sqrt{b} +5)} :\frac{5\sqrt{b} }{b-10\sqrt{b} +25} =\frac{5}{(\sqrt{b} -5)(\sqrt{b} +5)} \cdot\frac{(\sqrt{b}-5)^{2} }{5\sqrt{b} } =$

$\displaystyle\bf\\=\frac{\sqrt{b} -5}{\sqrt{b}\cdot(\sqrt{b} +5) } \\\\\\3)\\\\\frac{\sqrt{b} -5}{\sqrt{b} \cdot(\sqrt{b}+5) } -\frac{2}{\sqrt{b} +5} =\frac{\sqrt{b} -5-2\sqrt{b} }{\sqrt{b} \cdot(\sqrt{b} +5)} =-\frac{\sqrt{b} +5}{\sqrt{b}\cdot(\sqrt{b} +5) } =-\frac{1}{\sqrt{b} } \\\\\\b=\frac{16}{49} \\\\\\-\frac{1}{\sqrt{\frac{16}{49} } } =-\frac{1}{\frac{4}{7} } =-\frac{7}{4}=-1,75$