rozwiaż równanie !

[Blancos31]

a) \(\displaystyle{ x^{3}}\) - 3x + 2 = 0
b) \(\displaystyle{ 2x^{3}}\) - \(\displaystyle{ 3x^{2}}\) + 1 = 0
c) \(\displaystyle{ x^{3}}\) - \(\displaystyle{ 7x^{2}}\) + 36 = 0

[maise]

a)
\(\displaystyle{ x^3-3x+2=x^2(x-1)+x(x-1)-2(x-1)=(x-1)(x^2+x-2)=\\
=(x+1)(x(x-1)+2(x-1))=(x-1)(x-1)(x+2)=(x+2)(x-1)^2\\
\\
(x+2)(x-1)^2=0\\
\Rightarrow (x+2)=0 \vee (x-1)=0}\)



b)
\(\displaystyle{ 2x^3-3x^2+1=2x^2(x-1)-x(x-1)-1(x-1)=(x-1)(2x^2-x-1)=\\
=(x-1)(2x(x-1)+1(x-1))=(x-1)(x-1)(2x+1)=(x-1)^2(2x+1)\\
\\
(x-1)^2(2x+1)=0\\
\Rightarrow (x-1)=0 \vee (2x+1)=0\\}\)

-- 16 marca 2009, 21:30 --

c)
\(\displaystyle{ x^3-7x^2+36=x^2(x-6)-x(x-6)-6(x-6)=(x-6)(x^2-x-6)=\\
=(x-6)(x(x+2)-3(x+2))=(x-6)(x-3)(x+2)\\
\\
(x-6)(x-3)(x+2)=0\\
\Rightarrow (x-6)=0 \vee (x-3)=0 \vee (x+2)=0}\)