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\(\frac{xy^3-x^2y}{x^2+xy}=\frac{xy\left(y^2-x\right)}{x\left(x+y\right)}=\frac{y\left(y^2-x\right)}{x+y}=\frac{y^3-xy}{x+y}\)
Với đk trên ta có:
P = \(\frac{2}{x}-\left(\frac{x^2}{x^2+xy}+\frac{y^2-x^2}{xy}-\frac{y^2}{xy+y^2}\right).\frac{x+y}{x^2+xy+y^2}\)
\(=\frac{2}{x}-\left(\frac{x}{x+y}-\frac{\left(x-y\right)\left(x+y\right)}{xy}-\frac{y}{x+y}\right).\frac{x+y}{x^2+xy+y^2}\)
\(=\frac{2}{x}-\left(\frac{x-y}{x+y}-\frac{\left(x-y\right)\left(x+y\right)}{xy}\right).\frac{x+y}{x^2+xy+y^2}\)
\(=\frac{2}{x}-\frac{x-y}{xy}.\left(xy-\left(x+y\right)^2\right).\frac{1}{x^2+xy+y^2}\)
\(=\frac{2}{x}+\frac{x-y}{xy}\)
\(=\frac{x+y}{xy}\)
\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)
\(=\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)\left(x+y\right)^2}\)
\(=\frac{10y}{15\left(x+y\right)^2}\)
\(\frac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\frac{\left(x^2-x\right)-\left(xy-y\right)}{\left(x^2-x\right)+\left(xy-y\right)}\)
\(=\frac{x\left(x-1\right)-y\left(x-1\right)}{x\left(x-1\right)+y\left(x-1\right)}\)
\(=\frac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}\)
\(=\frac{x-y}{x+y}\)
a)\(\frac{2xy}{3\left(x+y\right)^2}\)
b)=\(\frac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)=\(\frac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}\)
=\(\frac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}\)=\(\frac{\left(x-y\right)}{\left(x+y\right)}\)
\(\dfrac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\dfrac{\left(x+y\right)\left(x+2y\right)}{x\left(x^2-y^2\right)+2y\left(x^2-y^2\right)}\)
\(=\dfrac{x+y}{x^2-y^2}\)
\(=\dfrac{1}{x-y}\)
A = x-y/x.(x+y) - 3x+y/x.(x-y) . (y-x)/x+y
= x-y/x.(x+y) + 3x+y/x.(x+y)
= x-y+3x+y/x.(x+y)
= 4x/x.(x+y)
= 4/x+y
Tk mk nha
\(A=\frac{x-y}{xy+y^2}-\frac{3x+y}{x^2-xy}.\frac{y-x}{x+y}\)
\(=\frac{x-y}{y\left(x+y\right)}-\frac{3x+y}{x\left(x-y\right)}.\frac{-\left(x-y\right)}{x+y}\)
\(=\frac{x-y}{y\left(x+y\right)}-\frac{-\left(3x+y\right).\left(x-y\right)}{x\left(x-y\right).\left(x-y\right)}\)
\(=\frac{x-y}{y\left(x+y\right)}-\frac{-\left(3x+y\right)}{x\left(x-y\right)}\)
\(=\frac{x\left(x-y\right)^2}{xy\left(x+y\right)\left(x-y\right)}+\frac{y\left(3x+y\right)\left(x+y\right)}{xy\left(x+y\right)\left(x-y\right)}\)
\(=\frac{x\left(x^2-2xy+y^2\right)+y\left(3x^2+4xy+y^2\right)}{xy\left(x^2-y^2\right)}\)
\(=\frac{x^4-2x^2y+xy^2+3x^2y+4xy^2+y^3}{xy\left(x^2-y^2\right)}\)
\(=\frac{x^4+x^2y+5xy^2+y^3}{xy\left(x^2-y^2\right)}=\frac{x^2\left(x^2+y\right)+y^2\left(5x+y\right)}{xy\left(x^2-y^2\right)}\)
\(A=x^2\left(x-y^2\right)-xy\left(1-xy\right)-x^3\\ =x^3-x^2y^2-xy+x^2y^2-x^3\\ =\left(x^3-x^3\right)+\left(-x^2y^2+x^2y^2\right)-xy\\ =-xy\)
\(\frac{xy^3-x^2}{x^2+xy}=\frac{x\left(y^3-x\right)}{x\left(x+y\right)}=\frac{y^3-x}{x+y}\)