Odpowiedź:
[tex]a)\\\\2x(5x^2-1)+2=(x-1)^2+2x(x+1)\\\\10x^3-2x+2=x^2-2x+1+2x^2+2x\\\\10x^3-2x+2=3x^2+1\\\\10x^3-2x+2-3x^2-1=0\\\\10x^3-3x^2-2x+1=0\\\\10x^3+5x^2-8x^2-4x+2x+1=0\\\\5x^2(2x+1)-4x(2x+1)+1(2x+1)=0\\\\(2x+1)(5x^2-4x+1)=0\\\\2x+1=0\ \ \ \ \ \ \ \ \vee\ \ \ \ 5x^2-4x+1=0\\\\2x=-1\ \ /:2\ \ \ \ \vee\ \ \ \ \Delta=b^2-4ac=(-4)^2-4*5*1=16-20=-4\\\\x=-\frac{1}{2}\ \ \ \ \ \ \ \ \ \ \ \ \vee\ \ \ \ \Delta<0\ \ \ \ -\ \ brak\ \ rozwiazania\\\\Odp.\ \ x=-\frac{1}{2}[/tex]
[tex]b)\\\\2(8x^3+1)+1=4x(7x-1)\\\\16x^3+2+1=28x^2-4x\\\\16x^3+3-28x^2+4x=0\\\\16x^3-28x^2+4x+3=0\\\\16x^3-8x^2-20x^2+10x-6x+3=0\\\\8x^2(2x-1)-10x(2x-1)-3(2x-1)=0\\\\(2x-1)(8x^2-10x-3)=0\\\\(2x-1)(8x^2+2x-12x-3)=0\\\\(2x-1)(2x(4x+1)-3(4x+1))=0\\\\(2x-1)(4x+1)(2x-3)=0\\\\2x-1=0\ \ \ \ \vee\ \ \ \ 4x+1=0\ \ \ \ \ \ \vee\ \ \ \ 2x-3=0\\\\2x=1\ \ /:2\ \ \vee\ \ \ \ 4x=-1\ \ /:4\ \ \vee\ \ \ \ 2x=3\ \ /:2\\\\x=\frac{1}{2}\ \ \ \ \ \ \ \ \ \ \vee\ \ \ \ x=-\frac{1}{4}\ \ \ \ \ \ \ \ \ \ \vee\ \ \ \ x=\frac{3}{2}[/tex]
