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a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)
c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)
Ta có
\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)= \(\frac{2y}{3\left(x+y\right)^2}\)
\(\frac{x^2+2x+1}{5x^3+5x^2}=\frac{\left(x+1\right)^2}{5x^2\left(x+1\right)}=\frac{x+1}{5x^2}\)
a)\(\frac{\left(x+1\right)^2}{5x^2\left(x+1\right)}\)=\(\frac{x+1}{5x^2}\)
b)\(\frac{10y}{15\left(x+y\right)^2}\)
Ta có
\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)= \(\frac{2y}{3\left(x+y\right)^2}\)
\(\frac{7x^2+14x+7}{3x^2+3x}=\frac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\frac{7\left(x+1\right)}{3x}\)
Với đk trên ta có:
P = \(\frac{2}{x}-\left(\frac{x^2}{x^2+xy}+\frac{y^2-x^2}{xy}-\frac{y^2}{xy+y^2}\right).\frac{x+y}{x^2+xy+y^2}\)
\(=\frac{2}{x}-\left(\frac{x}{x+y}-\frac{\left(x-y\right)\left(x+y\right)}{xy}-\frac{y}{x+y}\right).\frac{x+y}{x^2+xy+y^2}\)
\(=\frac{2}{x}-\left(\frac{x-y}{x+y}-\frac{\left(x-y\right)\left(x+y\right)}{xy}\right).\frac{x+y}{x^2+xy+y^2}\)
\(=\frac{2}{x}-\frac{x-y}{xy}.\left(xy-\left(x+y\right)^2\right).\frac{1}{x^2+xy+y^2}\)
\(=\frac{2}{x}+\frac{x-y}{xy}\)
\(=\frac{x+y}{xy}\)
\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)
\(=\frac{4y^2-\left(x-y\right)^2}{y^2\left(x-y\right)}.\frac{y^2-xy}{x-3y}+\frac{x\left(x-2y\right)-2\left(x^2-xy\right)}{2\left(x-2y\right)}.\frac{2x-4y}{xy+y^2}\)
\(=\frac{3y^2+2xy-x^2}{y^2\left(x-y\right)}.\frac{y^2-xy}{x-3y}+\frac{-x^2}{2\left(x-2y\right)}.\frac{2x-4y}{xy+y^2}\)
\(=\frac{\left(x+y\right)\left(3y-x\right)}{y^2\left(x-y\right)}.\frac{y\left(y-x\right)}{x-3y}-\frac{x^2}{2\left(x-2y\right)}.\frac{2\left(x-2y\right)}{y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)}{y}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}=\frac{2xy+y^2}{y\left(x+y\right)}=\frac{2x+y}{x+y}\)
Giờ chỉ cần thế x, y vô nữa là xong nhé.
\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)
\(=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y\left(y-x\right)}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x\left(x-y\right)}{x-2y}\right):\frac{y\left(x+y\right)}{2\left(x-2y\right)}\)
\(=\frac{4y\left(y-x\right)}{\left(x-y\right)\left(x-3y\right)}-\frac{\left(x-y\right)y\left(y-x\right)}{y^2\left(x-3y\right)}\)\(+\frac{x.2\left(x-2y\right)}{2.y\left(x+y\right)}-\frac{x\left(x-y\right).2\left(x-2y\right)}{\left(x-2y\right).y\left(x+y\right)}\)
\(=\frac{-4y}{x-3y}+\frac{\left(x-y\right)^2}{y\left(x-3y\right)}+\frac{x\left(x-2y\right)}{y\left(x+y\right)}-\frac{2x\left(x-y\right)}{y\left(x+y\right)}\)
\(=\frac{-4y^2+x^2-2xy+y^2}{y\left(x-3y\right)}+\frac{x^2-2xy-2x^2+2xy}{y\left(x+y\right)}\)
\(=\frac{x^2-2xy-3y^2}{y\left(x-3y\right)}+\frac{-x^2}{y\left(x+y\right)}\)
\(=\frac{x^2+xy-3xy-3y^2}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{x\left(x+y\right)-3y\left(x+y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(\frac{\left(x+y\right)\left(x-3y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{x+y}{y}-\frac{x^2}{y\left(x+y\right)}=\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}\)
\(=\frac{x^2-2xy+y^2-x^2}{y\left(x+y\right)}=\frac{-2xy+y^2}{y\left(x+y\right)}\)
\(=\frac{y\left(y-2x\right)}{y\left(x+y\right)}=\frac{y-2x}{x+y}\)
Thay \(x=\frac{1}{2};y=\frac{1}{3}\)vào A ta có :
\(A=\frac{\frac{1}{3}-2.\frac{1}{2}}{\frac{1}{2}+\frac{1}{3}}=\frac{\frac{1}{3}-1}{\frac{3}{6}+\frac{2}{6}}=\frac{2}{3}:\frac{5}{6}=\frac{2.6}{3.5}=\frac{4}{5}\)
Vậy \(A=\frac{4}{5}\)tại \(x=\frac{1}{2};y=\frac{1}{3}\)
\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)
\(=\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)\left(x+y\right)^2}\)
\(=\frac{10y}{15\left(x+y\right)^2}\)
\(\frac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\frac{\left(x^2-x\right)-\left(xy-y\right)}{\left(x^2-x\right)+\left(xy-y\right)}\)
\(=\frac{x\left(x-1\right)-y\left(x-1\right)}{x\left(x-1\right)+y\left(x-1\right)}\)
\(=\frac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}\)
\(=\frac{x-y}{x+y}\)
a)\(\frac{2xy}{3\left(x+y\right)^2}\)
b)=\(\frac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)=\(\frac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}\)
=\(\frac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}\)=\(\frac{\left(x-y\right)}{\left(x+y\right)}\)