Изрази и уравнения: (x^2+1)/(x) + (x)/(x^2+1) = 2,9

[lalala]

(2x^2 + √2x) /( 2x^2 + 2√2x+1)

(x^2+1)/(x) + (x)/(x^2+1) = 2,9

(4)/(x^2+4) + (5)/(x^2+5) = 2

[Xixibg]

[tex]1.\frac{2x^2+\sqrt{2}x}{2x^2+2\sqrt{2}+1}=\frac{x\cancel{(2x+\sqrt{2})}}{\cancel{(2x+\sqrt{2})}(x+\frac{\sqrt{2}}{2})}=\frac{2x}{2x+\sqrt{2}}[/tex]

[tex]2. \frac{x^2+1}{x}=y ; \frac{x}{x^2+1}=\frac{1}{y} ;y\ne0[/tex]
[tex]y+\frac{1}{y}=2,9[/tex]
[tex]y^2-2,9y+1=0[/tex]
[tex]D=8,41-4=4,41=2,1^2[/tex]
[tex]y_1=\frac{2,9-2,1}{2}=0,4 ; =>x^2+1=0,4x ; =>x^2-0,4x+1=0 ; D<0 =>[/tex] няма решение
[tex]y_2=\frac{2,9+2,1}{2}=2,5 ; =>x^2+1=2,5x ; =>x^-2,5x+1=0[/tex]
[tex]D=6,25-4=2,25=1,5^2[/tex]
[tex]x_1=\frac{2,5-1,5}{2}=0,5[/tex]
[tex]x_2=\frac{2,5+1,5}{2}=2[/tex]

[tex]3.\frac{4}{x^2+4}+\frac{5}{x^2+5}=2[/tex]
[tex]\frac{4}{x^2+4}\le 1[/tex]
[tex]\frac{5}{x^2+5}\le 1[/tex]
[tex]=>\frac{4}{x^2+4}+\frac{5}{x^2+5}\le 2 ; =>x=0[/tex]