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a/ \(ab-2b-3a+6=\left(ab-2b\right)-\left(3a-6\right)=b\left(a-2\right)-3\left(a-2\right)=\left(a-2\right)\left(b-3\right)\)
b/ \(ax-by-ay+bx==\left(ax+bx\right)-\left(by+ay\right)=x\left(a+b\right)-y\left(b+a\right)=\left(a+b\right)\left(x-y\right)\)
c/ \(ax+by-ay-bx=\left(ax-ay\right)+\left(by-bx\right)=a\left(x-y\right)+b\left(y-x\right)=a\left(x-y\right)-b\left(x-y\right)=\left(x-y\right)\left(a-b\right)\)
d/ \(a^2-\left(b+c\right)a+bc=a^2-ab-ac+bc=\left(a^2-ac\right)+\left(ab-bc\right)=a\left(a-c\right)+b\left(a-c\right)=\left(a-c\right)\left(a+b\right)\)e/ \(\left(3a-2\right)\left(4a-3\right)-\left(2-3a\right)\left(3a+1\right)=\left(3a-2\right)\left(4a-3\right)+\left(3a-2\right)\left(3a+1\right)=\left(3a-2\right)\left(4a-3+3a+1\right)=\left(3a-2\right)\left(7a-2\right)\)
f/ \(ax+ay+az-bx-by-bz-x-y-z=\left(ax+ay+az\right)-\left(bx+by+bz\right)-\left(x+y+z\right)\)
\(=a\left(x+y+z\right)-b\left(x+y+z\right)-\left(x+y+z\right)=\left(x+y+z\right)\left(a-b-1\right)\)
a/ \(x\left(a+b\right)+y\left(a+b\right)=\left(x+y\right)\left(a+b\right)\)
b/ \(a\left(x+y\right)+b\left(x+y\right)-1\left(x+y\right)=\left(a+b-1\right)\left(x+y\right)\)
c/ \(=x^2z\left(x+y-z-yz\right)\)
a/ \(ab+bd-ac-cd\)
\(=\left(ab+ac\right)-\left(bd+cd\right)\)
\(=a\left(b+c\right)-d\left(b+c\right)\)
\(=\left(b+c\right)\left(a-d\right)\)
b/ \(ax+by-ay-bx\)
\(=\left(ax-ay\right)-\left(bx-by\right)\)
\(=a\left(x-y\right)-b\left(x-y\right)\)
\(=\left(x-y\right)\left(a-b\right)\)
c/ \(x^2-xy-xy+y^2\)
\(=\left(x^2-xy\right)-\left(xy-y^2\right)\)
\(=x\left(x-y\right)-y\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y\right)\)
a) \(ab+bd-ac-cd\)
\(=\left(ab+bd\right)-\left(ac+cd\right)\)
\(=b\left(a+d\right)-c\left(a+d\right)\)
\(=\left(a+d\right)\left(b-c\right)\)
b) \(ax+by-ay-bx\)
\(=ax-bx+by-ay\)
\(=\left(ax-bx\right)-\left(ay-by\right)\)
\(=x\left(a-b\right)-y\left(a-b\right)\)
\(=\left(a-b\right)\left(x-y\right)\)
c) \(x^2-xy-xy+y^2\)
\(=\left(x^2-xy\right)-\left(xy-y^2\right)\)
\(=x\left(x-y\right)-y\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y\right)\)
\(=\left(x-y\right)^2\)
Từ hằng kết quả trên ta suy ra được hằng đẳng thức :
\(x^2-2xy+y^2\) :)
1. Ta có:
a) \(\left(x-2y\right)\left(3xy-2y+3x\right)\)
\(=x\left(3xy-2y+3x\right)-2y\left(3xy-2y+3x\right)\)
\(=3x^2y-2xy+3x^2-6xy^2+4y^2-6xy\)
\(=3x^2y-6xy^2+3x^2-8xy+4y^2\)
b) \(\left(x-1\right)\left(x-2\right)\left(x-3\right)=\left(x-1\right)\left[\left(x-2\right)\left(x-3\right)\right]\)
\(=\left(x-1\right)\left[x\left(x-3\right)-2\left(x-3\right)\right]\)
\(=\left(x-1\right)\left(x^2-3x-2x+6\right)\)
\(=\left(x-1\right)\left(x^2-5x+6\right)\)
\(=x\left(x^2-5x+6\right)-1\left(x^2-5x+6\right)\)
\(=x^3-5x^2+6x-x^2+5x-6\)
\(=x^3-6x^2+11x-6\)
a)\(ax-by+bx-ay\)
\(=\left(ax-ay\right)+\left(bx-by\right)\)
\(=a\left(x-y\right)+b\left(x-y\right)\)
\(=\left(a+b\right)\left(x-y\right)\)
b) sai đề