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a) \(3x^2-3y^2=3\left(x^2-y^2\right)=3\left(x-y\right)\left(x+y\right)\)
b) \(x^2-xy+7x-7y=\left(x^2+7x\right)-\left(xy+7y\right)\)
\(=x\left(x+7\right)-x\left(y+7\right)=x\left(x+7-y-7\right)=x\left(x-y\right)\)
c)\(x^2-3x+2=x^2-2x-x+2=\left(x^2-x\right)-\left(2x-2\right)\)
\(=x\left(x-1\right)-2\left(x-1\right)=\left(x-2\right)\left(x-1\right)\)
d) \(x^3+2x^2y+xy^2-16x=x\left(x^2+2xy+y^2-16\right)\)
\(=x\left[\left(x+y\right)^2-16\right]=x\left(x+y-4\right)\left(x+y+4\right)\)
a ) x ^ 2 + 2xy + 7x + 7y + y ^2 + 10 = ( x + y ) ^2 + 7 ( x + y ) + 10 = ( x + y ) ( x + y + 17 )
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a) \(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
b) \(x^2-25+y^2+2xy\)
\(=\left(x^2+2xy+y^2\right)-25\)
\(=\left(x+y\right)^2-5^2\)
\(=\left(x+y+5\right)\left(x+y-5\right)\)
a) \(ax+ay-3x-3y=a\left(x+y\right)-3\left(x+y\right)=\left(a-3\right)\left(x+y\right)\)
b) \(x^3-3x^2+3x-9=x^2\left(x-3\right)+3\left(x-3\right)=\left(x-3\right)\left(x^2+3\right)\)
c) xem lại đề
d) \(9-x^2-2xy-y^2=9-\left(x+y\right)^2=\left(3-x-y\right)\left(3+x+y\right)\)
\(a,A=x^2+7x+7y-y^2\\ =x^2-y^2+7x+7y\\ =\left(x-y\right)\left(x+y\right)+7\left(x+y\right)\\ =\left(x+y\right)\left(x-y+7\right)\)
\(b,B=x^2+2xy+y^2-3x-3y\\ =\left(x+y\right)^2-3\left(x+y\right)\\ =\left(x+y\right)\left(x+y-3\right)\)