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a) \(x\left(x^2-2x\right)+\left(x-2x\right)=x^2\left(x-2\right)+x\left(x-2\right)=\left(x-2\right)\left(x^2+x\right)⋮x-2\forall x,y\in Z\)
b) \(x^3y^2-3yx^2+xy=xy\left(x^2y-3x+1\right)⋮xy\forall x,y\in Z\)
c) \(x^3y^2-3x^2y^3+xy^2=xy^2\left(x^2-3xy+1\right)⋮\left(x^2-3xy+1\right)\forall x,y\in Z\)
Là khai triển đa thức hay tính hả em? Muốn tính thì phải có điều kiện của $x$ chứ?
a) 6x2 - 12x
= 6x(x - 2)
b) x2 + 2x + 1 - y2
= (x2 + 2x + 1) - y2
= (x + 1)2 - y2
= (x + 1 - y)(x + 1 + y)
c) x + y + z + x2 + xy + xz
= (x + x2) + (y + xy) + (z + xz)
= x(1 + x) + y(1 + x) + z(1 + x)
= (x + y + z)(x + 1)
d) xy + xz + y2 + yz
= (xy + xz) + (y2 + yz)
= x(y + z) + y(y + z)
= (x + y)(x + z)
e) x3 + x2 + x + 1
= (x3 + x2) + (x + 1)
= x2(x + 1) + (x + 1)
= (x2 + 1)(x + 1)
f) xy + y - 2x - 2
= (xy + y) - (2x + 2)
= y(x + 1) - 2(x + 1)
= (y - 2)(x + 1)
g) x3 + 3x - 3x2 - 9
= (x3 - 3x2) + (3x - 9)
= x2(x - 3) + 3(x - 3)
= (x2 + 3)(x - 3)
h) x2 - y2 - 2x - 2y
= (x2 - y2) - (2x + 2y)
= (x + y)(x - y) - 2(x + y)
= (x + y)(x - y - 2)
i) 7x2 - 7xy - 5x = 5y
mk thấy con này sai sai ý
đề bài là gì vậy bạn
Đề bài là tính
\(A,\left(x-y\right)\left(x^2+xy+y^2\right)+2y^3=x^3-y^3+2y^3=x^3+y^3\)
\(B,\frac{2x}{3x-3y}:\frac{x^2}{x-y}=\frac{2x}{3.\left(x-y\right)}.\frac{x-y}{x^2}=\frac{2x\left(x-y\right)}{3x^2\left(x-y\right)}=\frac{2}{3x}\)
\(C,\left(x-2\right)\left(\frac{3}{x+2}-\frac{5}{2x-4}+\frac{8}{x^2-4}\right)=\left(x-2\right)\left(\frac{3}{x+2}-\frac{5}{2.\left(x-2\right)}+\frac{8}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\left(x-2\right).\frac{6.\left(x-2\right)-5.\left(x+2\right)+8.2}{2.\left(x+2\right)\left(x-2\right)}=\left(x-2\right).\frac{6x-12-5x-10+16}{2.\left(x+2\right)\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x-6\right)}{2.\left(x+2\right)\left(x-2\right)}=\frac{x-6}{2.\left(x+2\right)}\)
\(D,\left(x-3\right)\left(\frac{2}{x+3}-\frac{3}{2x-6}+\frac{9}{x^2-9}\right)=\left(x-3\right)\left(\frac{2}{x+3}-\frac{3}{2.\left(x-3\right)}+\frac{9}{\left(x+3\right)\left(x-3\right)}\right)\)
\(=\left(x-3\right).\frac{4\left(x-3\right)-3\left(x+3\right)+9.2}{2.\left(x+3\right)\left(x-3\right)}=\left(x-3\right).\frac{4x-12-3x-9+18}{2\left(x+3\right)\left(x-3\right)}\)
\(=\frac{\left(x-3\right)\left(x-3\right)}{2.\left(x+3\right)\left(x-3\right)}=\frac{x-3}{2.\left(x+3\right)}\)