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\(x^5+x+1\)
\(=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
\(=x^2\left(x+y\right)-\left(x+y\right)=\left(x^2-1\right)\left(x+y\right)=\left(x-1\right)\left(x+1\right)\left(x+y\right)\)
\(x^2\left(x-3\right)-4x+12=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
=x²(x-3)-4x+3.4
=x²(x-3)-4(x+3)
=x²(x-3)+4(x-3)
=(x-3)(x²+4)
=(x-3)(x²+2²)
=(x-3)(x-2)(x+2)
\(x^3-2xy-x^2y+2y^2=\left(x^3-x^2y\right)-\left(2xy-2y^2\right)\)
\(=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x^2-2y\right)\left(x-y\right)\)
\(=x^2\left(x-y\right)-2y\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-2y\right)\)
Câu 1:
\(=x^2-\left(y-4\right)^2\)
\(=\left(x-y+4\right)\cdot\left(x+y-4\right)\)
\(x^4-81=\left(x^2-9\right)\left(x^2+9\right)=\left(x-3\right)\left(x+3\right)\left(x^2+9\right)\)
\(=\left(x^2-9\right)\left(x^2+9\right)=\left(x-3\right)\left(x+3\right)\left(x^2+9\right)\)
Bài 2:
Sửa đề: \(x^3-3x^2-10x=0\)
\(\Leftrightarrow x\left(x^2-3x-10\right)=0\)
\(\Leftrightarrow x\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-2\end{matrix}\right.\)
x^7+x^5+1=x^7+x^6+x^5-x^6+1
=x^5(x^2+x+1)-[(x^3)^2-1]
=x^5(x^2+x+1)-(x^3+1)(x^3-1)
=x^5(x^2+x+1)-(x^3+1)(x-1)(x^2+x+1)
=(x^2+x+1)[x^5-(x^3+1)(x-1)]
=(x^2+x+1)(x^5-x^4+x^3-x+1)