[tex]( \frac{x}{x+1})^2+( \frac{x+1}{x} )^2= \frac{17}{4} [/tex]

[x/(x+1)]²=aa+1/a=17/44a²-17a+4=0D=289-64=225a1=(17-15)/8=1/4⇒[x/(x+1)]²=1/4[x/(x+1)=-1/2⇒2x=-x-1,x≠-1⇒2x+x=-1⇒3x=-1⇒x=-1/3[x/(x+1)=1/2⇒2x=x+1,x≠-1⇒2x-x=1⇒x=1a2=(17+15)/8=4⇒[x/(x+1)]²=4[x/(x+1)=-2⇒x=-2x-2,x≠-1⇒x+2x=-2⇒3x=-2⇒x=-2/3[x/(x+1)=2⇒x=2x+2,x≠-1⇒2x-x=-2⇒x=-2Ответ x={-2;-2/3;-1/3;1}