x . ( x^3 + 3x^2+ x – 5 ) = 0 Help plz;—; 23/08/2021 Bởi Mackenzie x . ( x^3 + 3x^2+ x – 5 ) = 0 Help plz;—;
$x(x³+3x²+x-5)=0$ $⇒x(x³-x²+4x²-4x+5x-5)=0$ $⇒x[x²(x-1)+4x(x-1)+5(x-1)]=0$ $⇒x(x-1)(x²+4x+5)=0$ Vì $x²+4x+5>0 ∀x$ ⇒\(\left[ \begin{array}{l}x=0\\x-1=0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\) $@David2k5$ Bình luận
x . ( x^3 + 3x^2+ x – 5 ) = 0 ⇔x . (x−1)(x²+4x+5) = 0 vì x²+4x+5 > 0 nên x = 0 hoặc x – 1 = 0 ⇔ x = 1 vậy S={ 0 ; 1 } Bình luận
$x(x³+3x²+x-5)=0$
$⇒x(x³-x²+4x²-4x+5x-5)=0$
$⇒x[x²(x-1)+4x(x-1)+5(x-1)]=0$
$⇒x(x-1)(x²+4x+5)=0$
Vì $x²+4x+5>0 ∀x$
⇒\(\left[ \begin{array}{l}x=0\\x-1=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
$@David2k5$
x . ( x^3 + 3x^2+ x – 5 ) = 0
⇔x . (x−1)(x²+4x+5) = 0
vì x²+4x+5 > 0
nên x = 0
hoặc x – 1 = 0 ⇔ x = 1
vậy S={ 0 ; 1 }