Question

Помогите решить задание на прикрепленном фото

Answer 1

$3.;(2-sqrt5)sqrt{9+4sqrt5}=frac{(2-sqrt5)sqrt{9+4sqrt5}sqrt{9+4sqrt5}}{sqrt{9+4sqrt5}}=frac{(2-sqrt5)({9+4sqrt5})}{sqrt{9+4sqrt5}}=\=frac{18+8sqrt5-9sqrt5-20}{sqrt{9+4sqrt5}}=frac{-2-sqrt5}{sqrt{9+4sqrt5}}=-frac{2+sqrt5}{sqrt{9+4sqrt5}}=t\t^2=left(-frac{2+sqrt5}{sqrt{9+4sqrt5}}right)^2=frac{(2+sqrt5)^2}{(sqrt{9+4sqrt5})^2}=frac{4+4sqrt5+5}{9+4sqrt5}=frac{9+4sqrt5}{9+4sqrt5}=1\t=sqrt{t^2}=sqrt1=pm1$


$4.;begin{cases}frac6{x+y}+frac5{x-y}=7\frac3{x+y}-frac2{x-y}=1end{cases}Rightarrowbegin{cases}2left(1+frac2{x-y}right)+frac5{x-y}=7\frac3{x+y}=1+frac2{x-y}end{cases}\2left(1+frac2{x-y}right)+frac5{x-y}=7\2+frac4{x-y}+frac5{x-y}=7\frac9{x-y}=5\x-y=frac95\x=1,8+y\begin{cases}x=1,8+y\frac3{1,8+y+y}=1+frac2{1,8+y-y}end{cases}$
$frac3{1,8+y+y}=1+frac2{1,8+y-y}\frac{3}{1,8+2y}=1+frac2{1,8}\frac3{1,8+2y}=frac{19}9\19cdot(1,8+2y)=3cdot9\34,2+38y=27\38y=-7,2\y=-frac{7,2}{38}=-frac{18}{95}\begin{cases}x=frac{153}{95}\y=-frac{18}{95}end{cases}$