Bài 8: Phân tích đa thức thành nhân tử bằng phương pháp nhóm các hạng tử

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a/x^3+3x^2+3x+2=0

b/x^4-2x^3+2x-1=0

c/x^4-3x^3-6x^2+8x=0

Khôi Bùi
27 tháng 8 2018 lúc 20:49

a ) \(x^3+3x^2+3x+2=0\)

\(\Leftrightarrow x^3+3x^2+3x+1+1=0\)

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

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

\(\Leftrightarrow x+1=-1\)

\(\Leftrightarrow x=-2\)

Vậy \(x=-2\)

b ) \(x^4-2x^3+2x-1=0\)

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

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

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

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

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

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

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

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

trần thị trâm anh
27 tháng 8 2018 lúc 20:55

a, \(x^3+3x^2+3x+2=0\)

\(\Leftrightarrow\left(x^3+2x^2\right)+\left(x^2+2x\right)+\left(x+2\right)=0\)

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

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

b, \(x^4-2x^3+2x-1=0\)

\(\Leftrightarrow\left(x^4-x^3\right)-\left(x^3-x^2\right)-\left(x^2-x\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x^3-x^2\right)\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x=1\)

đỗ việt
13 tháng 12 2020 lúc 19:30

a ) x^3+3x^2+3x+2=0x3+3x2+3x+2=0

\Leftrightarrow x^3+3x^2+3x+1+1=0⇔x3+3x2+3x+1+1=0

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

 

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

\Leftrightarrow x+1=-1⇔x+1=−1

\Leftrightarrow x=-2⇔x=−2

Vậy x=-2x=−2

b ) x^4-2x^3+2x-1=0x4−2x3+2x−1=0

\Leftrightarrow x^4-1-2x\left(x^2-1\right)=0⇔x4−1−2x(x2−1)=0

\Leftrightarrow\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)=0⇔(x2−1)(x2+1)−2x(x2−1)=0

\Leftrightarrow\left(x^2-1\right)\left(x^2+1-2x\right)=0⇔(x2−1)(x2+1−2x)=0

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

\Leftrightarrow\left(x-1\right)^3\left(x+1\right)=0⇔(x−1)3(x+1)=0

\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x=-1\end{matrix}\right.⇔[(x−1)3=0x+1=0​⇔[x−1=0x=−1​

\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.⇔[x=1x=−1​

 

Vậy \left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.[x=1x=−1​  

Hậuu
23 tháng 12 2020 lúc 5:45

a ) x^3+3x^2+3x+2=0x3+3x2+3x+2=0

\Leftrightarrow x^3+3x^2+3x+1+1=0⇔x3+3x2+3x+1+1=0

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

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

\Leftrightarrow x+1=-1⇔x+1=−1

\Leftrightarrow x=-2⇔x=−2

Vậy x=-2x=−2

b ) x^4-2x^3+2x-1=0x4−2x3+2x−1=0

\Leftrightarrow x^4-1-2x\left(x^2-1\right)=0⇔x4−1−2x(x2−1)=0

\Leftrightarrow\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)=0⇔(x2−1)(x2+1)−2x(x2−1)=0

\Leftrightarrow\left(x^2-1\right)\left(x^2+1-2x\right)=0⇔(x2−1)(x2+1−2x)=0

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

\Leftrightarrow\left(x-1\right)^3\left(x+1\right)=0⇔(x−1)3(x+1)=0

\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x=-1\end{matrix}\right.⇔[(x−1)3=0x+1=0​⇔[x−1=0x=−1​

\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.⇔[x=1x=−1​

Vậy \left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.[x=1x=−1​ 

Nguyễn Quốc Trung
27 tháng 12 2020 lúc 21:35

a ) x3+3x2+3x+2=0x3+3x2+3x+2=0

⇔x3+3x2+3x+1+1=0⇔x3+3x2+3x+1+1=0

⇔(x+1)3+1=0⇔(x+1)3+1=0

⇔(x+1)3=−1⇔(x+1)3=−1

⇔x+1=−1⇔x+1=−1

⇔x=−2⇔x=−2

Vậy x=−2x=−2

b ) x4−2x3+2x−1=0x4−2x3+2x−1=0

⇔x4−1−2x(x2−1)=0⇔x4−1−2x(x2−1)=0

⇔(x2−1)(x2+1)−2x(x2−1)=0⇔(x2−1)(x2+1)−2x(x2−1)=0

⇔(x2−1)(x2+1−2x)=0⇔(x2−1)(x2+1−2x)=0

⇔(x−1)(x+1)(x−1)2=0⇔(x−1)(x+1)(x−1)2=0

⇔(x−1)3(x+1)=0⇔(x−1)3(x+1)=0

⇔[(x−1)3=0x+1=0⇔[x−1=0x=−1⇔[(x−1)3=0x+1=0⇔[x−1=0x=−1

⇔[x=1x=−1⇔[x=1x=−1

Vậy [x=1x=−1


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