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a) \(x^3+x^2y-x^2z-xyz\)
\(=x^2\left(x+y\right)-xz\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xz\right)\)
\(=x\left(x+y\right)\left(x-z\right)\)
b) \(x^2-6x+9-9y^2\)
\(=\left(x^2-2\cdot x\cdot3+3^2\right)-\left(3y\right)^2\)
\(=\left(x-3\right)^2-\left(3y\right)^2\)
\(=\left(x-3-3y\right)\left(x-3+3y\right)\)
c) \(x^2+9x+20\)
\(=x^2+5x+4x+20\)
\(=x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(x+5\right)\left(x+4\right)\)
d) \(x^4+4\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot2+4-2\cdot x^2\cdot2\)
\(=\left(x^2+2\right)-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
a/\(x^3+x^2y-x^2z-xyz\)
\(=\left(x^3-x^2y\right)+\left(x^2y-xyz\right)\)
\(=x^2\left(x-z\right)+xy\left(x-z\right)\)
\(=\left(x-z\right)\left(x^2+xy\right)\)
b/\(x^2-6x+9-9y^2\)
\(=\left(x^2-6x+9\right)-9y^2\)
\(=\left(x-3\right)^2-\left(3y\right)^2\)
\(=\left(x-3+3y\right)\left(x-3-3y\right)\)
c/\(x^2+9x+20\)
\(=x^2+4x+5x+20\)
\(=\left(x^2+4x\right)+\left(5x+20\right)\)
\(=x\left(x+4\right)+5\left(x+4\right)\)
\(=\left(x+5\right)\left(x+4\right)\)
d/\(x^4+4\)
\(=x^4+4x^2-4x^2+4\)
\(=\left(x^2+4x^2+4\right)-4x^2\)
\(=\left(x+2\right)^2-\left(2x\right)^2\)
\(=\left(x+2-2x\right)\left(x+2+2x\right)\)
\(x^3-x^2y+3x-3y\)
\(=x^2\left(x-y\right)+3\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+3\right)\)
\(=x^2\left(x-y\right)+3\left(x-y\right)=\left(x^2+3\right)\left(x-y\right)\)
Câu 1:
\(=x^2-\left(y-4\right)^2\)
\(=\left(x-y+4\right)\cdot\left(x+y-4\right)\)
\(x^4+x^3+2x^2+x+1=\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\\ =x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+1\right)\left(x^2+x+1\right)\)
Dễ thấy \(x^2+1>0\); \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) nên ta không thể phân tích thêm được nữa.
Vậy \(x^4+x^3+2x^2+x+1=\left(x^2+1\right)\left(x^2+x+1\right)\).
\(x^3\left(2+x\right)^2-\left(x+2\right)^2+1-x^3\\ =\left(x+2\right)^2\left(x^3-1\right)-\left(x^3-1\right)\\ =\left[\left(x+2\right)^2-1\right]\left(x^3-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x^2+4x+3\right)=\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x^2+x+1\right)\)
\(x^2\left(x-3\right)-4x+12=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
=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)
\(x^3-2xy-x^2y+2y^2=\left(x^3-x^2y\right)-\left(2xy-2y^2\right)\)
\(=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x^2-2y\right)\left(x-y\right)\)
\(=x^2\left(x-y\right)-2y\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-2y\right)\)
\(x^5+x+1\)
\(=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
