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a: \(=\left(3-x\right)\left(x+1\right)\)
b: \(=3x\left(x-y\right)-5\left(x-y\right)\)
=(x-y)(3x-5)
c: \(=x\left(x-y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)\left(x-10\right)\)
a) \(=x\left(3-x\right)+\left(3-x\right)=\left(3-x\right)\left(x+3\right)\)
b) \(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
c) \(=x\left(x-y\right)-10\left(x-y\right)=\left(x-y\right)\left(x-10\right)\)
d) \(=\left(x+y\right)^2-16=\left(x+y-4\right)\left(x+y+4\right)\)
e) \(=\left(x-y\right)\left(x+y\right)-4\left(x+y\right)=\left(x+y\right)\left(x-y-4\right)\)
f) \(=9-\left(4x^2-4xy+y^2\right)=9-\left(2x-y\right)^2=\left(3-2x+y\right)\left(3+2x-y\right)\)
g) \(=y\left(y^2-2xy+x^2-y\right)\)
h) \(=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
i) \(=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left(2x+y\right)\)
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9: \(\left(-2x\right)\left(3x^2-2x+4\right)=-6x^3+4x^2-8x\)
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`#3107.101107`
`N = A - B`
`N = -3x^3y^2 - x^2y + 3xy - 1 - (-x^2y - 3x^3y^2 + 3xy - 3)`
`= -3x^3y^2 - x^2y + 3xy - 1 + x^2y + 3x^3y^2 - 3xy + 3`
`= (-3x^3y^2 + 3x^3y^2) + (-x^2y + x^2y) + (3xy - 3xy) + (-1 + 3)`
`= 2`
Bậc của đa thức N (?) là `0.`
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a: \(\dfrac{\left(x+1\right)}{x^2+2x-3}=\dfrac{\left(x+1\right)}{\left(x+3\right)\cdot\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+5\right)}{\left(x+3\right)\left(x-1\right)\left(x+2\right)\left(x+5\right)}\)
\(\dfrac{-2x}{x^2+7x+10}=\dfrac{-2x}{\left(x+2\right)\left(x+5\right)}=\dfrac{-2x\left(x+3\right)\left(x-1\right)}{\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x-1\right)}\)
b: \(\dfrac{x-y}{x^2+xy}=\dfrac{x-y}{x\left(x+y\right)}=\dfrac{y^2\left(x-y\right)}{xy^2\left(x+y\right)}\)
\(\dfrac{2x-3y}{xy^2}=\dfrac{\left(2x-3y\right)\left(x+y\right)}{xy^2\left(x+y\right)}\)
c: \(\dfrac{x-2y}{2}=\dfrac{\left(x-2y\right)\left(x-xy\right)}{2\left(x-xy\right)}\)
\(\dfrac{x^2+y^2}{2x-2xy}=\dfrac{x^2+y^2}{2\left(x-xy\right)}\)
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\(VT=\dfrac{2x^2+2xy+xy+y^2}{x^2\left(2x+y\right)-y^2\left(2x+y\right)}=\dfrac{2x\left(x+y\right)+y\left(x+y\right)}{\left(x^2-y^2\right)\left(2x+y\right)}\\ =\dfrac{\left(2x+y\right)\left(x+y\right)}{\left(2x+y\right)\left(x-y\right)\left(x+y\right)}=\dfrac{1}{x-y}=VP\)
\(\left(3xy-x^2+y\right).\frac{2}{3}x^2y\)-
= \(3xy.\)\(\frac{2}{3}x^2y\)\(-\) \(x^2.\frac{2}{3}x^2y\)\(+\)\(y.\frac{2}{3}x^2y\)
= \(x^3y^2\)\(-\)\(\frac{2}{3}x^4y\)\(+\)\(\frac{2}{3}x^2y^2\)