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Thiếu y3 nha bạn :
\(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
\(x^3-3xy^2-2y^3\)
\(=x^3-xy^2-2xy^2-2y^3\)
\(=x\left(x^2-y^2\right)-2y^2\left(x+y\right)\)
\(=x\left(x-y\right)\left(x+y\right)-2y^2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy-2y^2\right)\)
\(=\left(x+y\right)\left(x^2-2xy+xy-2y^2\right)\)
\(=\left(x+y\right)\left[x\left(x-2y\right)+y\left(x-2y\right)\right]\)
\(=\left(x+y\right)^2\left(x-2y\right)\)
\(x^3-3xy^2-2y^3\)
\(=x^3-xy^2-2xy^2-2y^3\)
\(=x\left(x^2-y^2\right)-2y^2\left(x+y\right)\)
\(=x\left(x-y\right)\left(x+y\right)-2y^2\left(x+y\right)\)
\(=\left(x+y\right)\left[x\left(x+y\right)-2y^2\right]\)
\(=\left(x+y\right)\left(x^2+xy-2y^2\right)\)
\(=\left(x+y\right)\left(x^2+2xy-xy-2y^2\right)\)
\(=\left(x+y\right)\left[x\left(x-2y\right)-y\left(x-2y\right)\right]\)
\(=\left(x-y\right)\left(x-y\right)\left(x-2y\right)\)
\(=\left(x-y\right)^2\left(x-2y\right)\)
c/ Ta có:
\(x^2-3xy+x-3y\)
\(=x^2+x-3xy-3y\)
\(=x\left(x+1\right)-3y\left(x+1\right)\)
\(=\left(x+1\right)\left(x-3y\right)\)
d/ Ta có:
\(x^3-x^2-5x+125\)
\(=x^3+5x^2-6x^2-30x+25x+125\)
\(=x^2\left(x+5\right)-6x\left(x+5\right)+25\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^2-3xy+x-3y\)
\(=x\left(x-3y\right)+\left(x-3y\right)\)
\(=\left(x+1\right)\left(x-3y\right)\)
\(x^3-x^2-5x+125\) k có nghiệm
\(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right).\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1-3xy\right]\)
\(=\left(x+y-1\right).\left[x^2+2xy+y^2+x+y+1-3xy\right]\)
\(=\left(x+y-1\right)\left(x^2+y^2-xy+x+y+1\right)\)
Chúc bạn học tốt.
\(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)
\(\Leftrightarrow\left(x+y-1\right)\left(\left(x+y\right)^2+\left(x+y\right).1+1^2\right)-3xy\left(x+y-1\right)\)
\(\Leftrightarrow\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)
\(\Leftrightarrow\left(x+y-1\right)\left(x^2-xy+y^2+x+y+1\right)\)
\(x^3+3xy+y^3-1=\left(x^3+y^3\right)+\left(x^2+2xy+y^2-1\right)-\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x+y-1\right)\left(x+y+1\right)-\left(x^2-xy+y^2\right)\)
\(=\left(x^2-xy+y^2\right)\left(x+y-1\right)+\left(x+y-1\right)\left(x+y+1\right)\)
\(=\left(x+y-1\right)\left(x^2-xy+y^2+x+y+1\right)\)
\(x^3+3xy+y^3-1\)
\(=\left(x+y\right)^3-1-3x^2y-3xy^2+3xy\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+y^2+2xy+x+y+1-3xy\right)\)
\(=\left(x+y-1\right)\left(x^2+y^2-xy+x+y+1\right)\)
(x+y)3- 1 - 3xy(x+y-1)
=(x+y)3 -13 -3xy(x+y-1)
=(x+y-1)3 -3xy(x+y-1)
=(x+y-1) [(x+y-1)-3xy]
=(x+y-1) [x+y-1-3xy]
chúc bạn học tốt nha!
\(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)
\(=x^3+3x^2y+3xy^2+1-1-3x^2y-3xy^2+3xy\)
\(=x^3+3xy\)
\(=x\left(x^2+3y\right)\)
3x2 - x - 3xy + y
= 3x(x-y) - (x-y)
= (3x-1)(x-y)