giải phương trình tìm x

T

thaolovely1412

Ta có: [TEX]x^3-x^2-8x+12=x^3+3x^2-4x^2-12x+4x+12=x^2(x+3)-4x(x+3)+4(x+3)=(x^2-4x+4)(x+3)=(x-2)^2(x+3)[/TEX]
\Rightarrow [TEX](x-2)^2(x+3)=0[/TEX]
\Leftrightarrow [TEX]\left[\begin{(x-2)^2=0}\\{x+3 = 0} [/TEX]
\Leftrightarrow [TEX]\left[\begin{x=2}\\{x=-3} [/TEX]
 
T

thienbinhgirl

$x^3+2x^2-x^2-2x-6x-12
=x^2(x+2)-x(x+2)-6(x+2)
=(x+2)(x^2-x-6)
=(x+2)(x^2-3x+2x-6)
=(x+2)(x+2)(x-3)
=(x+2)^2(x-3)$
\Rightarrow hoặc $(x+2)^2$=0\Rightarrow x=-2 ;
hoặc $(x-3)$=0\Rightarrow x=3
 
Last edited by a moderator:
H

howare

câu trả lời

x^3-x^2-8x+12=0
\Leftrightarrow x^3+3x^2-4x^2-12x+4x+12=0
\Leftrightarrow (x+3)(x^2-4x+4)=0
\Leftrightarrow (x+3)(x-2)^2=0
\Leftrightarrow x=-3;x=2
 
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