Question

Помогите 20 баллов !!!!!

Answer 1

$\displaystyle \frac{9x-y}{3x+x^{0.5}y^{0.5}} = \frac{(3x^{0.5}-y^{0.5})(3x^{0.5}+y^{0.5})}{x^{0.5}(3x^{0.5}+y^{0.5})} = \frac{3x^{0.5}-y^{0.5}}{x^{0.5}} = \frac{3x- \sqrt{xy} }{x}$

Если x=10 и y=576, то $\displaystyle \frac{3x- \sqrt{xy} }{x} = \dfrac{3\cdot 10-\sqrt{10\cdot 576}}{10} = \frac{30- \sqrt{5760} }{10}$


$\displaystyle \frac{4x-y}{2x+x^{0.5}y^{0.5}} = \frac{(2x^{0.5}-y^{0.5})(2x^{0.5}+y^{0.5})}{x^{0.5}(2x^{0.5}+y^{0.5})}= \frac{2x^{0.5}-y^{0.5}}{x^{0.5}}= \frac{2x- \sqrt{xy} }{x}$

Если x=32 и y=81, то $\displaystyle \frac{2x- \sqrt{xy} }{x} =\frac{2\cdot 32- \sqrt{32\cdot81} }{32} =2- \frac{9\cdot 4\sqrt{2}}{32} =2- \frac{9\sqrt{2}}{8}$

Answer 2

B44.
$\frac{9x-y}{3x+x^{0,5}*y^{0,5}} = \frac{(3x^{0,5})^2-(y^{0,5})^2}{x^{0,5}(3x^{0,5}+y^{0,5})} = \\ \\ = \frac{(3x^{0,5}-y^{0,5})*(3x^{0,5}+y^{0,5})}{x^{0,5}(3x^{0,5}+y^{0,5})} =\frac{3x^{0,5}-y^{0,5}}{x^{0,5}} = \\ \\ =3 - \frac{y^{0,5}}{x^{0,5}} =3- \sqrt{ \frac{y}{x} }$

x = 10; y = 576
$3- \sqrt{ \frac{y}{x} } =3- \sqrt{\frac{576}{10}} =3- \frac{24}{ \sqrt{10} } =3-2,4 \sqrt{10}$

B45.
$\frac{4x-y}{2x+x^{0,5}y^{0,5}} = \frac{(2 \sqrt{x} )^2-( \sqrt{y} )^2}{2 (\sqrt{x})^2 + \sqrt{xy} } = \\ \\ =\frac{(2 \sqrt{x}- \sqrt{y} )(2 \sqrt{x}+ \sqrt{y})}{\sqrt{x}(2 \sqrt{x} + \sqrt{y} ) } =\frac{2 \sqrt{x}- \sqrt{y}}{\sqrt{x} } =2- \sqrt{\frac{y}{x}}$

x = 32;  y = 81
$2- \sqrt{\frac{y}{x}} =2- \sqrt{\frac{81}{32}} =2- \frac{9}{4 \sqrt{2} } =2- \frac{9 \sqrt{2} }{8} =2-1,125 \sqrt{2}$