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

PN

Tìm x

a,x3-4x2-9x+36=0

b,5x2-4(x2-2x+1)-5=0

c,(x2-9)2-(x-3)=0

d,x3-3x+2=0

MP
21 tháng 9 2017 lúc 18:47

a) \(x^3-4x^2-9x+36=0\Leftrightarrow x^3-7x^2+12x+3x^2-21x+36=0\) \(x\left(x^2-7x+12\right)+3\left(x^2-7x+12\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-7x+12\right)=0\) \(\Leftrightarrow\left(x+3\right)\left(x^2-7x+12\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x-4x+12\right)=0\) \(\Leftrightarrow\left(x+3\right)\left(x\left(x-3\right)-4\left(x-3\right)\right)=0\Leftrightarrow\left(x+3\right)\left(x-4\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\x-4=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=4\\x=3\end{matrix}\right.\) vậy \(x=-3;x=4;x=3\)

b) \(5x^2-4\left(x^2-2x+1\right)-5=0\) \(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\Leftrightarrow x^2-x+9x-9=0\)

\(\Leftrightarrow x\left(x-1\right)+9\left(x-1\right)\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+9=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\) vậy \(x=-9;x=1\)

c) đề có sai o bn

d) \(x^3-3x+2=0\Leftrightarrow x^3+x^2-2x-x^2-x+2=0\)

\(\Leftrightarrow x\left(x^2+x-2\right)-\left(x^2+x-2\right)=0\Leftrightarrow\left(x-1\right)\left(x^2+x-2\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-x+2x-2\right)=0\Leftrightarrow\left(x-1\right)\left(x\left(x-1\right)+2\left(x-1\right)\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x-1\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x+2=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-2\\x=1\end{matrix}\right.\)

vậy \(x=1;x=-2\)

Bình luận (5)
LV
21 tháng 9 2017 lúc 18:26

1. \(x^3-4x^2-9x+36=0\)

\(\Rightarrow x^2.\left(x-4\right)-9\left(x-4\right)=0\)

\(\Rightarrow\left(x^2-9\right)\left(x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2-9=0\Rightarrow x\in\left\{3;-3\right\}\\x-4=0\Rightarrow x=4\end{matrix}\right.\)

Vậy ..........

2. \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Rightarrow5x^2-4\left(x^2-1\right)-5=0\)

\(\Rightarrow5x^2-4x^2+4-5=0\)

\(\Rightarrow x^2-1=0\)

\(\Rightarrow x^2=1\)

\(\Rightarrow x=\pm1\)

Vậy .......

3. \(x^3-3x+2=0\)

\(\Rightarrow x^3-4x+x+2=0\)

\(\Rightarrow x.\left(x^2-4\right)+x+2=0\)

\(\Rightarrow x.\left(x-2\right).\left(x+2\right)+x+2=0\)

\(\Rightarrow\left(x+2\right).\left(x^2-2x+1\right)=0\)

\(\Rightarrow\left(x+2\right)\left(x-1\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x+2\right)=0\\\left(x-1\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

Vậy .......

Bình luận (1)