Phân tích đa thức thành nhân tử:
x^2 - x - 6
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\(x^2\left(x-3\right)-4x+12=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
=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)
\(x^3-2xy-x^2y+2y^2=\left(x^3-x^2y\right)-\left(2xy-2y^2\right)\)
\(=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x^2-2y\right)\left(x-y\right)\)
\(=x^2\left(x-y\right)-2y\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-2y\right)\)
\(=x^2\left(x+y\right)-\left(x+y\right)=\left(x^2-1\right)\left(x+y\right)=\left(x-1\right)\left(x+1\right)\left(x+y\right)\)
\(x^5+x+1\)
\(=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
Bài 2:
Sửa đề: \(x^3-3x^2-10x=0\)
\(\Leftrightarrow x\left(x^2-3x-10\right)=0\)
\(\Leftrightarrow x\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-2\end{matrix}\right.\)
\(x-y-\sqrt{x}-\sqrt{y}\\ =x-y-\left(\sqrt{x}+\sqrt{y}\right)\\ =\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}+\sqrt{y}\right)\\ =\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}-1\right)\)
=(x-y)-(căn x+căn y)
=(căn x-căn y)(căn x+căn y)-(căn x+căn y)
=(căn x+căn y)(căn x-căn y-1)
\(=\left(x\sqrt{x}+y\sqrt{y}\right)+\left(x-y\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(x+y-\sqrt{xy}+\sqrt{x}-\sqrt{y}\right)\)
Có : x^2-x-6 = (x^2-3x)+(2x-6) = x.(x-3)+2.(x-3) = (x-3).(x+2)
Tk mk nha
\(=x^2+2x-3x-6\)
\(=x.\left(x+2\right)-3.\left(x+2\right)\)
\(=\left(x-3\right).\left(x+2\right)\)