1/Phân tíchcác đa thức sau: a) x+2a(x-y)-y b) 5ay – 3bc + ac -15by 2/Tìm x biết: a)x(x-2) + x-2=0 b)x^3 +x^2 + x+1=0 c)2(x+3)-x^2 -3x=0 d)2x(3x-5) =1

Đáp án:

1/ a/ $(x-y)(1+2a)$

b/ $(a-3b)(c+5y)$

2/ a/ $x=2$ hoặc $x=-1$

b/ $x=-1$

c/ $x=-3$ hoặc $x=2$

d/ $x=\dfrac{5}{3}$ hoặc $x=-2$

Giải thích các bước giải:

1/ a/ $x+2a(x-y)-y=(x-y)+2a(x-y)=(x-y)(1+2a)$

b/ $5ay-3bc+ac-15by=c(a-3b)+5y(a-3b)=(a-3b)(c+5y)$

2/ a/ $x(x-2)+x-2=0$

$⇔ (x-2)(x+1)=0$

$⇔ \left[ \begin{array}{l}x-2=0\\x+1=0\end{array} \right.$

$⇔ \left[ \begin{array}{l}x=2\\x=-1\end{array} \right.$

b/ $x^3+x^2+x+1=0$

$⇔ x^2(x+1)+(x+1)=0$

$⇔ (x+1)(x^2+1)=0$

$\text{Vì $x^2+1 \geq 1 > 0$}$

$\text{nên $x+1=0 ⇔ x=-1$}$

c/ $2(x+3)-x^2-3x=0$

$⇔ 2(x+3)-x(x+3)=0$

$⇔ (x+3)(2-x)=0$

$⇔ \left[ \begin{array}{l}x+3=0\\2-x=0\end{array} \right.$

$⇔ \left[ \begin{array}{l}x=-3\\x=2\end{array} \right.$

d/ $2x(3x-5)=10-6x$

$⇔ 2x(3x-5)-10+6x=0$

$⇔ 2x(3x-5)+2(3x-5)=0$

$⇔ 2(3x-5)(x+2)=0$

$⇔ \left[ \begin{array}{l}3x-5=0\\x+2=0\end{array} \right.$

$⇔ \left[ \begin{array}{l}x=\dfrac{5}{3}\\x=-2\end{array} \right.$