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\(x^6+2x^5+x^4-2x^3-2x^2+1=\left(x^3+x^2-1\right)^2\)
a) \(=\left(x^2-6\right)\left(x^2-1\right)=\left(x^2-6\right)\left(x-1\right)\left(x+1\right)\)
b) \(=\left(x^2-1\right)\left(x^2+3\right)=\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\)
c) \(=x^2\left(x-1\right)-x\left(x-1\right)+4\left(x-1\right)=\left(x-1\right)\left(x^2-x+4\right)\)
#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^6-x^4+2x^3+2x^2\)
\(=x^2\left(x^4-x^2+2x+2\right)\)
\(=x^2\left[x^4-2x^3+x^2+2x^3-4x^2+2x+2x^2-4x+2\right]\)
\(=x^2\left[x^2\left(x^2-2x+1\right)+2x\left(x^2-2x+1\right)+2\left(x^2-2x+1\right)\right]\)
\(=x^2\left(x^2-2x+1\right)\left(x^2+2x+2\right)\)
\(=x^2\left(x-1\right)^2\left(x^2+2x+2\right)\)
\(x^6-x^4+2x^3+2x\)
\(=x^5x-x^3x+2x^2x+2x\)
\(=x\left(x^5-x^3+2x^2+2\right)\)
Bài 1:
b: \(3x-6=x^2-16\)
\(\Leftrightarrow x^2-3x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
\(x^6-x^4+2x^3+2x^2=x^2\left(x^4-x^2+2x+2\right)\)
\(=x^2\left(\left(x^4+2x^3+x^2\right)+\left(-2x^3-4x^2-2x\right)+\left(2x^2+4x+2\right)\right)\)
\(=x^2\left(x^2+2x+1\right)\left(x^2-2x+2\right)\)
\(=x^2\left(x+1\right)^2\left(x^2-2x+2\right)\)
xin lỗi nhưng mình chưa hiểu lắm, bannj có thể viết rõ hơn không