Question

(1+y^2)^2=4y(1+y)^2
^2-в квадрате

Answer 1

$(1+y^2)^2=4y(1+y)^2;\y^4+2y^2+1=4y(y^2+2y+1);\y^4+2y^2+1=4y^3+8y^2+4y;\y^4-4y^3-6y^2-4y+1=0 |:y^2neq 0;\y^2-4y-6-frac{4}{y}+frac{1}{y^2}=0;\y^2+frac{1}{y^2}-4(y+frac{1}{y})-6=0;\(y+frac{1}{y})^2-2-4(y+frac{1}{y})-6=0;\(y+frac{1}{y})^2-4(y+frac{1}{y})-8=0.$

$t=y+frac{1}{y}.\t^2-4t-8=0;\D_t/4=4+8=12;\t_1=2-2sqrt{3},\t_2=2+2sqrt{3}.$

$y+frac{1}{y}=2-2sqrt{3};\f(y)=y+frac{1}{y},\E(f)=(-infty; -2]U[2; +infty];\f(y)neq2-2sqrt{3}.$

$y+frac{1}{y}=2+2sqrt{3};\ y^2-(2+2sqrt{3})y+1=0;\D_y/4=(1+sqrt{3})^2-1=1+2sqrt{3}+3-1=3+2sqrt{3};\y_1=1+sqrt{3}+sqrt{3+2sqrt{3}};\y_2=1+sqrt{3}-sqrt{3+2sqrt{3}}.$