# Phân tích đa thức thành nhân tử bằng cách phối hợp nhiều phương pháp

Tìm x, biết:

a) x. (2x - 7) - 4x + 14 = 0

b) x2. (x - 1) - 4x + 4 = 0

c) x + x2 - x3 - x4 = 0

d) 2x3 + 3x2 + 2x + 3 = 0

e) 4x2 - 25 - (2x - 5). (2x + 7) = 0

g) x3 + 27 + (x + 3). (x - 9) = 0

Lương Minh Hằng 3 tháng 8 2019 lúc 15:58

$x\left(2x-7\right)-4x+14=0\Leftrightarrow\left(x-2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{2}\end{matrix}\right.$

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

$x^4-x^3-x^2+x=x\left(x^3+1\right)-x^2\left(x+1\right)=x\left(x+1\right)\left(x^2-x+1-x^2\right)=x\left(x+1\right)\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.$

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a) $x\left(2x-7\right)-4x+14-0\Leftrightarrow2x^2-11x+14=0\Leftrightarrow2x^2-4x-7x+14=0\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=2\end{matrix}\right.$

b) $x^2\left(x-1\right)-4x+4=0\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.$

c) $x+x^2-x^3-x^4=0\Leftrightarrow x\left(x^3+x^2-x-1\right)=0\Leftrightarrow x\left[x\left(x^2-1\right)+\left(x^2-1\right)\right]=0\Leftrightarrow x\left(x+1\right)\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.$

d) $2x^3+3x^2+2x+3=0\Leftrightarrow x^2\left(2x+3\right)+2x+3=0\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\Leftrightarrow x=-1,5\left(x^2+1>0\forall x\right)$

e) $4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\Leftrightarrow2x-5=0\Leftrightarrow x=2,5$

g) $x^3+27+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.$

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Duyên 3 tháng 8 2019 lúc 16:19

a) x. (2x - 7) - 4x + 14 = 0

⇔ 2x$^2$ - 7x - 4x + 14 =0

⇔ 2x( x - 2 ) - 7 ( x - 2 ) = 0

⇔ ( 2x - 7 ) ( x - 2 ) = 0

$\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=2\end{matrix}\right.$

b) x2. (x - 1) - 4x + 4 = 0

⇔ x2. (x - 1) - 4( x - 1 ) = 0

⇔(x$^2$ - 4 ) ( x - 1 ) = 0

$\Leftrightarrow\left[{}\begin{matrix}x^2-4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm2\\x=1\end{matrix}\right.$

d) 2x3 + 3x2 + 2x + 3 = 0

⇔ 2x( x$^2$ + 1 ) +3( x$^2$ + 1 ) = 0

⇔ ( 2x + 3 ) ( x$^2$ + 1 ) = 0

$\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-3}{2}\\x=\pm1\end{matrix}\right.$

e) 4x2 - 25 - (2x - 5). (2x + 7) = 0

⇔ ( 2x - 5 ) ( 2x + 5 ) - ( 2x - 5 ) (2x + 7 ) = 0

⇔ ( 2x - 5 ) ( 2x + 5 - 2x - 7 ) = 0

⇔-2(2x - 5 ) =0

$\Leftrightarrow\left[{}\begin{matrix}-2=0\left(vl\right)\\2x-5=0\end{matrix}\right.$

⇔ x= $\frac{5}{2}$

g) x3 + 27 + (x + 3). (x - 9) = 0

⇔ ( x+ 3 ) ( x$^2$ - 3x + 9) + ( x+ 3 ) ( x - 9 ) = 0

⇔ ( x + 3 ) ( x$^2$ - 3x + 9 + x - 9 ) = 0

⇔ ( x + 3 ) ( x$^2$ - 2x ) = 0

$\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=0\end{matrix}\right.$

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