Question

Помогите, пожалуйста. Нужно подробное решение каждого номера.

Answer 1

1
(4-√6)²-(4+√6)²=(4-√6-4-√6)(4-√6+4+√6)=(-2√6)*8=-16√6
2
$x^2=( \sqrt{7+2 \sqrt{6} } - \sqrt{7-2 \sqrt{6} } )^2=7+2 \sqrt{6} -2 \sqrt{(7+2 \sqrt{6})(7-2 \sqrt{6} ) }+$$7-2 \sqrt{6} =14-2 \sqrt{49-24} =14-2* \sqrt{25} =14-2*5=14-10=4$
x=2 U x=-2
3
1)x/(x+y)-xy/(x+y)(y-x)=(xy-x²-xy)/(x+y)(y-x)=-x²/(x+y)(y-x)=x²(x+y)(x-y)
2)x²/(x-y)² : x²/(x-y)(x+y)=x²/(x-y)² * (x-y)(x+y)/x²=(x+y)/(x-y)

Answer 2

$(4-\sqrt6)^2-(4+\sqrt6)^2=16-8\sqrt6+6-16-8\sqrt6-6=-16\sqrt6$

$(\sqrt{7+2\sqrt6}-\sqrt{7-2\sqrt6})^2=7+2\sqrt6+7-2\sqrt6-\\-2\sqrt{(7+2\sqrt6)(7-2\sqrt6)}=14-2\sqrt{(49-24)}=14-2*5=4\\x^2=4\\x=^+_-2$

$\frac{x^2}{x^2-2xy+y^2}:(\frac{x}{x+y}-\frac{xy}{y^2-x^2})=\frac{x^2}{(x-y)^2}:\frac{xy-x^2-xy}{(y-x)(y+x)}=\frac{x^2}{(x-y)^2}*\\*(-\frac{(y-x)(y+x)}{x^2})=-\frac{y+x}{y-x}=\frac{x+y}{x-y}$