Bài 9: Phân tích đa thức thành nhân tử bằng cách phối hợp nhiều phương pháp

a: \(x^3-4x^2-x+4=0\)

=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)

=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)

=>\(\left(x+3\right)\left(x^2-4\right)=0\)

=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)

=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

Sửa đề

x² - 25y² - 6x + 9

= (x² - 6x + 9) - 25y²

= (x - 3)² - (5y)²

= (x - 3 - 5y)(x - 3 + 5y)

= (x - 5y - 3)(x + 5y - 3)

Từ điều kiện đã cho không tính được $x+y$ bạn nhé. Bạn xem lại đề.

a: \(\dfrac{x^2+3x}{x+3}=\dfrac{x\left(x+3\right)}{x+3}=x\)

b: \(\dfrac{x-2}{x^2-5x+6}=\dfrac{x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x-3}\)

c: \(\dfrac{1}{x+y}+\dfrac{y}{x^2-y^2}\)

\(=\dfrac{x-y+y}{x^2-y^2}=\dfrac{x}{x^2-y^2}\)

\(\left(a\right)\dfrac{x^2+3x}{x+3}=\dfrac{x\left(x+3\right)}{x+3}=x\\ \left(b\right)\dfrac{x-2}{x^2-5x+6}=\dfrac{x-2}{x^2-2x-3x+6}=\dfrac{x-2}{x\left(x-2\right)-3\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{x-3}\\ \left(c\right)\dfrac{1}{x+y}+\dfrac{y}{x^2-y^2}=\dfrac{x-y+y}{x^2-y^2}=\dfrac{x}{x^2-y^2}\)

\(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1^2\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x+1\right)\left(x-1\right)\)

\(=\left[\left(x-1\right)\left(x^2+x+1\right)\right]\left[\left(x+1\right)\left(x^2-x+1\right)\right]\)

\(=\left(x^3-1^3\right)\left(x^3+1^3\right)\)

\(=\left(x^3+1\right)\left(x^3-1\right)\)

\(=\left(x^3\right)^2-1^2\)

\(=x^6-1\)

Lời giải:
a. $x^3-4x^2+x+6=(x^3-2x^2)-(2x^2-4x)-(3x-6)$

$=x^2(x-2)-2x(x-2)-3(x-2)=(x-2)(x^2-2x-3)$
$=(x-2)[(x^2+x)-(3x+3)]=(x-2)[x(x+1)-3(x+1)]$

$=(x-2)(x+1)(x-3)$

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b.

$x^3+7x^2+14x+8=(x^3+x^2)+(6x^2+6x)+(8x+8)$

$=x^2(x+1)+6x(x+1)+8(x+1)=(x+1)(x^2+6x+8)$

$=(x+1)[(x^2+2x)+(4x+8)]=(x+1)[x(x+2)+4(x+2)]$

$=(x+1)(x+2)(x+4)$

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+3x+2x+6\right)\left(x^2+5x+4x+20\right)-24\)

\(=\left(x^2+5x+6\right)\left(x^2+9x+20\right)-24\)

\(=\left(x^4+9x^3+20x^2+5x^3+45x^2+100x+6x^2+54x+120\right)-24\)

\(=x^4+14x^3+71x^2+154x+120-24\)

\(=x^4+14x^3+71x^2+154x+96\)

\(=x^4+13x^3+58x^2+96x+x^3+13x^2+58x+96\)

\(=x\left(x^3+13x^2+58x+96\right)+\left(x^3+13x^2+58x+96\right)\)

\(=\left(x^3+13x^2+58x+96\right)\left(x+1\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)