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Đề Bài : Phân tích các đa thức sau thành nhân tử. Giúp E Với ạ 🥲

Nguyễn Lê Phước Thịnh · Accepted Answer

o: x^4+x^3+x^2-1=x^3(x+1)+(x-1)(x+1)=(x+1)(x^3+x-1)q: $=\left(x^3-y^3\right)+xy\left(x-y\right)$=(x-y)(x^2+xy+y^2)+xy(x-y)=(x-y)(x^2+2xy+y^2)=(x-y)(x+y)^2s: =(2xy)^2-(x^2+y^2-1)^2=(2xy-x^2-y^2+1)(2xy+x^2+y^2-1)=[1-(x^2-2xy+y^2]+[(x+y)^2-1]=(1-x+y)(1+x-y)(x+y-1)(x+y+1)u: =(x^2-y^2)-4(x+y)=(x+y)(x-y)-4(x+y)=(x+y)(x-y-4)x: =(x^3-y^3)-(3x-3y)=(x-y)(x^2+xy+y^2)-3(x-y)=(x-y)(x^2+xy+y^2-3)z: =3(x-y)+(x^2-2xy+y^2)=3(x-y)+(x-y)^2=(x-y)(x-y+3)Câu trả lời: o: x^4+x^3+x^2-1=x^3(x+1)+(x-1)(x+1)=(x+1)(x^3+x-1)q: $=\left(x^3-y^3\right)+xy\left(x-y\right)$=(x-y)(x^2+xy+y^2)+xy(x-y)=(x-y)(x^2+2xy+y^2)=(x-y)(x+y)^2s: =(2xy)^2-(x^2+y^2-1)^2=(2xy-x^2-y^2+1)(2xy+x^2+y^2-1)=[1-(x^2-2xy+y^2]+[(x+y)^2-1]=(1-x+y)(1+x-y)(x+y-1)(x+y+1)u: =(x^2-y^2)-4(x+y)=(x+y)(x-y)-4(x+y)=(x+y)(x-y-4)x: =(x^3-y^3)-(3x-3y)=(x-y)(x^2+xy+y^2)-3(x-y)=(x-y)(x^2+xy+y^2-3)z: =3(x-y)+(x^2-2xy+y^2)=3(x-y)+(x-y)^2=(x-y)(x-y+3)

HT.Phong (9A5) · Answer

o) $x^4+x^3+x^2-1$$=\left(x^4+x^3\right)+\left(x^2-1\right)$$=x^3\left(x+1\right)+\left(x+1\right)\left(x-1\right)$$=\left(x+1\right)\left(x^3+x-1\right)$q) $x^3+x^2y-xy^2-y^3$$=\left(x^3+x^2y\right)-\left(xy^2+y^3\right)$$=x^2\left(x+y\right)-y^2\left(x+y\right)$$=\left(x+y\right)\left(x^2-y^2\right)$$=\left(x+y\right)^2\left(x-y\right)$s) $4x^2y^2-\left(x^2+y^2-1\right)^2$$=\left(2xy\right)^2-\left(x^2+y^2-1\right)^2$$=\left(2xy-x^2-y^2+1\right)\left(2xy+x^2+y^2-1\right)$$=-\left(x^2-2xy+y^2-1\right)\left(x^2+2xy+y^2-1\right)$$=-\left(x-y-1\right)\left(x-y+1\right)\left(x+y+1\right)\left(x+y-1\right)$u) $x^2-y^2-4x-4y$$=\left(x^2-y^2\right)-\left(4x+4y\right)$$=\left(x+y\right)\left(x-y\right)-4\left(x+y\right)$$=\left(x+y\right)\left(x-y-4\right)$x) $x^3-y^3-3x+3y$$=\left(x^3-y^3\right)-\left(3x-3y\right)$$=\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x-y\right)$$=\left(x-y\right)\left(x^2+xy+y^2-3\right)$z) $3x-3y+x^2-2xy+y^2$$=\left(3x-3y\right)+\left(x^2-2xy+y^2\right)$$=3\left(x-y\right)+\left(x-y\right)^2$$=\left(x-y\right)\left(3+x-y\right)$Câu trả lời: o) $x^4+x^3+x^2-1$$=\left(x^4+x^3\right)+\left(x^2-1\right)$$=x^3\left(x+1\right)+\left(x+1\right)\left(x-1\right)$$=\left(x+1\right)\left(x^3+x-1\right)$q) $x^3+x^2y-xy^2-y^3$$=\left(x^3+x^2y\right)-\left(xy^2+y^3\right)$$=x^2\left(x+y\right)-y^2\left(x+y\right)$$=\left(x+y\right)\left(x^2-y^2\right)$$=\left(x+y\right)^2\left(x-y\right)$s) $4x^2y^2-\left(x^2+y^2-1\right)^2$$=\left(2xy\right)^2-\left(x^2+y^2-1\right)^2$$=\left(2xy-x^2-y^2+1\right)\left(2xy+x^2+y^2-1\right)$$=-\left(x^2-2xy+y^2-1\right)\left(x^2+2xy+y^2-1\right)$$=-\left(x-y-1\right)\left(x-y+1\right)\left(x+y+1\right)\left(x+y-1\right)$u) $x^2-y^2-4x-4y$$=\left(x^2-y^2\right)-\left(4x+4y\right)$$=\left(x+y\right)\left(x-y\right)-4\left(x+y\right)$$=\left(x+y\right)\left(x-y-4\right)$x) $x^3-y^3-3x+3y$$=\left(x^3-y^3\right)-\left(3x-3y\right)$$=\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x-y\right)$$=\left(x-y\right)\left(x^2+xy+y^2-3\right)$z) $3x-3y+x^2-2xy+y^2$$=\left(3x-3y\right)+\left(x^2-2xy+y^2\right)$$=3\left(x-y\right)+\left(x-y\right)^2$$=\left(x-y\right)\left(3+x-y\right)$

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