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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)
\(27x^6-8x^3=x^3\left(27x^3-8\right)=x^3\left[\left(3x\right)^3-2^3\right]=x^3\left(3x-2\right)\left(9x^2+6x+4\right)\)
\(x^6-y^6=\left(x^2\right)^3-\left(y^2\right)^3=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x-y\right)\left(x+y\right)\left[\left(x^2+y^2\right)^2-x^2y^2\right]\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2+xy\right)\left(x^2+y^2-xy\right)\)
\(y^9-9x^2y^6+27x^4y^3-27x^6=\left(y^3\right)^3-3.\left(y^3\right)^2.\left(3x^2\right)+3.y^3.\left(3x^2\right)^2-\left(3x^2\right)^3\)
\(=\left(y^3-3x^2\right)^3\)
a) Ta có : a2x + a2y - 7x - 7y
= a2(x + y) - (7x + 7y)
= a2(x + y) - 7(x + y)
= (x + y)(a2 - 7)
b) Ta có : x3 + y(1 - 3x2) + x(3x2 - 1) - y3
= x3 - y(3x2 - 1) + x(3x2 - 1) - y3
= x3 - y3 + [x(3x2 - 1) - y(3x2 - 1)]
= x3 - y3 - (3x2 - 1)(x - y)
= (x - y)(x2 + xy + y2) - (3x2 - 1)(x - y)
= (x - y)[(x2 + xy + y2) - (3x2 - 1)]
= (x - y)(x2 + xy + y2 - 3x2 + 1)
= (x - y)(-2x2 + xy + y2 + 1)
bài 2:a. \(5x.\left(y^2-2yz+z^2\right)\)
\(=5x.\left(y-z\right)^2\) .......k bít dc chưa
b.\(\left(x^2y-x\right)+\left(xy^2-y\right)\)
\(=x.\left(xy-1\right)+y.\left(xy-1\right)\)
\(=\left(xy-1\right).\left(x+y\right)\)
x2-y2+6x+6y = (x2-y2)+(6x+6y) = (x-y)(x+y)+6(x+y) = (x-y-6)(x+y)
\(5x^2-x+y-5y^2\)
\(=\left(5x^2-5y^2\right)-\left(x-y\right)\)
\(=5\left(x^2-y^2\right)-\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left[5\left(x+y\right)-1\right]\)
\(=\left(x-y\right)\left(5x+5y-1\right)\)
\(-3xy^2+x^2y^2-5x^2y\)
\(=-xy\left(3y+xy-5x\right)\)
\(x\left(y-1\right)+3\left(y^3+2y+1\right)\)
\(=3y^3+6y+3+xy-x\)
Xem lại nhé ko phân tích được
\(12xy^2-12xy+3x\)
\(=3x\left(4y^2-4y+1\right)\)
\(=3x\left(2y-1\right)^2\)
\(10x^2\left(x+y\right)-5\left(2x+2y\right)y^2\)
\(=10x^2\left(x+y\right)-10\left(x+y\right)y^2\)
\(=10\left(x+y\right)\left(x-y\right)\left(x+y\right)\)
\(=10\left(x+y\right)^2\left(x-y\right)\)
ab(x2+y2)+xy(a2+b2)
\(=abx^2+aby^2+a^2xy+b^2xy=\left(abx^2+a^2xy\right)+\left(aby^2+b^2xy\right).\)
\(=ax\left(bx+ay\right)+by\left(ay+bx\right)=\left(ax+by\right).\left(ay+bx\right)\)
\(2y^2-7y+3=2y\left(y-3\right)-\left(y-3\right)=\left(y-3\right)\left(2y-1\right)\)
\(y^3+y^2+y=y\left(y^2+y+1\right)\)
\(15y^2+19y+6=5y\left(3y+2\right)+3\left(3y+2\right)=\left(3y+2\right)\left(5y+3\right)\)
a) \(2y^2-7y+3=2y^2-6y-y+3\)
\(=2y\left(y-3\right)-\left(y-3\right)=\left(y-3\right)\left(2y-1\right)\)
b) \(y^3+y^2+y=y\left(y^2+y+1\right)\)
c) \(15y^2+19y+6=15y^2+10y+9y+6\)
\(=5y\left(3y+2\right)+3\left(3y+2\right)=\left(3y+2\right)\left(5y+3\right)\)