Question

Сократите дробь. Прошу помогите

Answer 1

$1) \ \dfrac{\sqrt{7}+7 }{\sqrt{7} }= \dfrac{\sqrt{7}+(\sqrt{7})^{2}}{\sqrt{7} }= \dfrac{\sqrt{7}\cdot(1+\sqrt{7})}{\sqrt{7} } =\boxed{1+\sqrt{7}}\\\\\\2) \ \dfrac{15-\sqrt{15} }{\sqrt{15} }=\dfrac{(\sqrt{15})^{2} -\sqrt{15} }{\sqrt{15} }=\dfrac{\sqrt{15}\cdot(\sqrt{15} -1) }{\sqrt{15} }=\boxed{\sqrt{15}-1}$

$3) \ \dfrac{(1-\sqrt{7} )^{2} }{\sqrt{7}-4 }=\dfrac{1-2\sqrt{7} +7}{\sqrt{7}-4 } =\dfrac{8-2\sqrt{7} }{\sqrt{7}-4 } =\dfrac{2\cdot(4-\sqrt{7}) }{\sqrt{7}-4 }=\boxed{-2}\\\\\\4) \ \dfrac{(1+\sqrt{2})^{2}}{\sqrt{\sqrt{3}+2\sqrt{2}} }=\dfrac{(1+\sqrt{2})^{2}}{\sqrt{1+2\sqrt{2}+2 } } =\dfrac{(1+\sqrt{2})^{2}}{\sqrt{(1+\sqrt{2} )^{2} } } =\dfrac{(1+\sqrt{2})^{2}}{1+\sqrt{2} }=\boxed{1+\sqrt{2}}$

Answer 2

Ответ:

Объяснение:

а) $(\sqrt{7} + 7) /\sqrt{7} = (\sqrt{7} + \sqrt{7} * \sqrt{7} ) /\sqrt{7} = (\sqrt{7} * (1 + (\sqrt{7}))/\sqrt{7} = 1 + \sqrt{7}$

б) $(15 - \sqrt{15} )/\sqrt{15} = (\sqrt{15} * \sqrt{15} - \sqrt{15})/\sqrt{15} = (\sqrt{15}*(\sqrt{15} - 1))/\sqrt{15} = \sqrt{15} - 1$

в) $(1 - \sqrt{7} )^{2} /(\sqrt{7} - 4) = (1 - 2\sqrt{7} + 7)/(\sqrt{7} - 4) = (8 - 2\sqrt{7})/(\sqrt{7} - 4) = -2(\sqrt{7} - 4)/(\sqrt{7} - 4) = -2$

г) $(1 + \sqrt{2})^{2} /(\sqrt{3 + 2\sqrt{2}}) = (1 + \sqrt{2} )^{2}/(\sqrt{1 + 2\sqrt{2} + 2 }) = (1 + \sqrt{2})^{2}/(\sqrt{(1 + \sqrt{2} )^{2}} = (1 + \sqrt{2})*(1 + \sqrt{2})/(1 + \sqrt{2}) = 1 + \sqrt{2}$