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x2-y2+6x+6y = (x2-y2)+(6x+6y) = (x-y)(x+y)+6(x+y) = (x-y-6)(x+y)
\(=6x^2+9x+4x+6\)
\(=3x.\left(2x+3\right)+2.\left(2x+3\right)\)
\(=\left(2x+3\right).\left(3x+2\right)\)
\(\)
\(4\left(x+3y-4\right)^2-x^2+6x-9\)
\(=\left[2\left(x+3y-4\right)\right]^2-\left(x^2-6x+9\right)\)
\(=\left[2x+6y-8\right]^2-\left(x-3\right)^2\)
\(=\left(2x+6y-8+x-3\right)\left(2x+6y-8-x+3\right)\)
\(=\left(3x+6y-11\right)\left(x+6y-5\right)\)
\(27x^6-8x^3=x^3\left(27x^3-8\right)=x^3\left[\left(3x\right)^3-2^3\right]=x^3\left(3x-2\right)\left(9x^2+6x+4\right)\)
\(x^6-y^6=\left(x^2\right)^3-\left(y^2\right)^3=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x-y\right)\left(x+y\right)\left[\left(x^2+y^2\right)^2-x^2y^2\right]\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2+xy\right)\left(x^2+y^2-xy\right)\)
\(y^9-9x^2y^6+27x^4y^3-27x^6=\left(y^3\right)^3-3.\left(y^3\right)^2.\left(3x^2\right)+3.y^3.\left(3x^2\right)^2-\left(3x^2\right)^3\)
\(=\left(y^3-3x^2\right)^3\)
\(2y^2-7y+3=2y\left(y-3\right)-\left(y-3\right)=\left(y-3\right)\left(2y-1\right)\)
\(y^3+y^2+y=y\left(y^2+y+1\right)\)
\(15y^2+19y+6=5y\left(3y+2\right)+3\left(3y+2\right)=\left(3y+2\right)\left(5y+3\right)\)
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)
\(5x^2-x+y-5y^2\)
\(=\left(5x^2-5y^2\right)-\left(x-y\right)\)
\(=5\left(x^2-y^2\right)-\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left[5\left(x+y\right)-1\right]\)
\(=\left(x-y\right)\left(5x+5y-1\right)\)
\(=\left(x+3\right)^6-y^6\\ =\left[\left(x+3\right)^3-y^3\right]\left[\left(x+3\right)^3+y^3\right]\\ =\left(x+3-y\right)\left[\left(x+3\right)^2+y\left(x+3\right)+y^2\right]\left(x+3+y\right)\left[\left(x+3\right)^2-y\left(x+3\right)+y^2\right]\\ =\left(x+y+3\right)\left(x-y+3\right)\left(x^2+6x+9+xy+3y+y^2\right)\left(x^2+6x+9-xy-3y+y^2\right)\)
\(\left(x^2+6x+9\right)^3-\left(y^2\right)^3=\left(x^2+6x+9-y^2\right)\left[\left(x^2+6x+9\right)^2+\left(x^2+6x+9\right)y^2+y^4\right]\)
\(=\left[\left(x+3\right)^2-y^2\right]\left\{\left[\left(x^2+6x+9\right)^2+2\left(x^2+6x+9\right)y^2+y^4\right]-\left(x^2+6x+9\right)y^2\right\}\)
\(=\left(x+3-y\right)\left(x+3+y\right)\left[\left(x^2+6x+9+y^2\right)^2-\left(x+3\right)^2y^2\right]\)
\(=\left(x+3-y\right)\left(x+3+y\right)\left[\left(x^2+6x+9+y^2\right)-\left(x+3\right)y\right]\left(x^2+6x+9+y^2\right)+\left(x+3\right)y\)
\(=\left(x+3-y\right)\left(x+3+y\right)\left(x^2+6x+9+y^2-xy-3y\right)\left(x^2+6x+9+y^2+xy+3y\right)\)