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\(x^2y-y+xy^2-x\)
=> \(x\left(xy-1\right)+y\left(-1+xy\right)\)
=> \(\left(-1+xy\right)\left(x+y\right)\)
a,\(x^2\)- xy - 8x + 8y
= \(\left(x^2-8x\right)\)- (xy - 8y)
= x( x - 8 ) - y( x - 8)
= (x - y)(x - 8)
a, \(x^8+x^7+1\)
= \(x^7\left(x+1\right)+1\)
= \(x^7\left(x+1\right)+1+x-x\)
= \(x^7\left(x+1\right)+\left(x+1\right)-x\)
= \(\left(x^7+1\right)\left(x+1\right)-x\)
a) \(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\) \(=x^6\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
b) \(x^4+64\)
\(=\left(x^2+8\right)^2-16x^2\)
\(=\left(x^2+8+4x\right)\left(x^2+8-4x\right)\)
2x2 + 10x + 8 = 2x2 + 2x + 8x + 8 = (2x2 + 2x) + (8x + 8)
= 2x(x + 1) + 8(x + 1) = (x + 1)(2x + 8)
bài này 1h rùi,chắc chờ tui ngủ dậy làm;
= (x+y)3 - (x+y) + xy(x+y) =
= (x+y)((x+y)2 -1 +xy)) = (x+y)(x2 +3xy +y2 -1)
a) x2 - 2xy - 4 + y2
= (x - y)2 - 22
= (x - y - 2)(x - y + 2)
b) x2 + y2 - 1 - 2xy
= (x - y)2 - 12
= (x - y - 1)(x - y + 1)
c) 25 - x2 + 4xy - 4y2
= 52 - (x - 2y)2
= (5 - x + 2y)(5 + x - 2y)
\(x^2+3y^2-4xy+10x-12y+9\)
\(=\left(x^2-xy+x\right)+9x-3xy+3y^2-12y+9\)
\(=\left(x^2-xy+x\right)+\left(9x-9y+9\right)-3xy+3y^2-3y\)
\(=\left(x^2-xy+x\right)+\left(9x-9y+9\right)-\left(3xy-3y^2+3y\right)\)
\(=x\left(x-y+1\right)+9\left(x-y+1\right)-3y\left(x-y+1\right)\\ =\left(x-y+1\right)\left(x+9-3y\right)\)
\(9-4x^2-4xy-y^2\)
⇔ \(3^2-\left[\left(2x\right)^2+2.2x.y+y^2\right]\)
⇔\(3^2-\left(2x+y\right)^2\)
⇔\(\left(3-2x+y\right)\left(3+2x+y\right)\)
\(9-4x^2-4xy-y^2\)
=> \(9-\left(2x+y\right)\left(2x+y\right)\)
=> \(-\left(2x+y+3\right)\left(2x+y+3\right)\)