Question

$( \frac{x-1}{x} )^{2} +( \frac{x-1}{x-2} )^{2}= \frac{40}{9}$

Answer 1

$\frac{ (x-1)^{2} }{ x^{2} } + \frac{(x-1)^{2}}{(x-2)^{2}} = \frac{40}{9}$
$\frac{(x-1)^{2}(x-2)^{2} + x^{2} (x-1)^{2}}{ x^{2} (x-2)^{2}} = \frac{40}{9}$
$\frac{(2x^{2}-4x+4)( x^{2}-2x+1)}{ x^{2}(x-2)^{2} } = \frac{40}{9}$
$\frac{2(x^{2}-2x+2)( x^{2}-2x+1)}{ (x^{2}-2x)^{2} } = \frac{40}{9}$ 
Замена x^2 - 2x = y. И делим все на 2
$\frac{(y+2)(y+1)}{ y^{2}} = \frac{20}{9}$
$9( y^{2}+3y+2) = 20 y^{2}$
$11 y^{2} -27y-18=0$
$(y-3)(11y+6)=0$
$y1 = x^{2} -2x=3; (x+1)(x-3)=0$
x1 = -1;x2 = 3
$y2= x^{2} -2x=-6/11; 11 x^{2} -22x+6=0$
$D/4= 11^{2} -11*6=121-66=55$
$x3=(11- \sqrt{55} )/11 = 1- \sqrt{ \frac{5}{11} }$
 $x3=(11+ \sqrt{55} )/11 = 1+ \sqrt{ \frac{5}{11} }$