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\(=\dfrac{2}{x}-\left(\dfrac{x^2}{x\left(x+y\right)}-\dfrac{x^2-y^2}{xy}-\dfrac{y^2}{y\left(x+y\right)}\right):\dfrac{x^3-y^3}{x^2-y^2}\)
\(=\dfrac{2}{x}-\left(\dfrac{x^2y-\left(x^2-y^2\right)\left(x+y\right)-y^2x}{xy\left(x+y\right)}\right)\cdot\dfrac{x+y}{x^2+xy+y^2}\)
\(=\dfrac{2}{x}-\dfrac{x^2y-x^3-x^2y+xy^2+y^3-xy^2}{xy}\cdot\dfrac{1}{x^2+xy+y^2}\)
\(=\dfrac{2}{x}-\dfrac{-\left(x-y\right)\left(x^2+xy+y^2\right)}{xy}\cdot\dfrac{1}{x^2+xy+y^2}\)
\(=\dfrac{2}{x}+\dfrac{x-y}{xy}=\dfrac{y+x-y}{xy}=\dfrac{1}{y}\)
\(A=\dfrac{\left(a+b\right)\left(-x-y\right)-\left(a-y\right)\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{a\left(-x-y\right)+b\left(-x-y\right)-a\left(b-x\right)+y\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-ax-ay-bx-by-ab+ax+by-xy}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-ay-bx-ab-xy}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-xy+ay+ab+by}{abxy\left(xy+ay+ab+by\right)}=\dfrac{-1}{abxy}\)
Với \(a=\dfrac{1}{3};b=-2;x=\dfrac{3}{2};y=1\)
\(\Rightarrow A=\dfrac{-1}{\dfrac{1}{3}.\left(-2\right).\dfrac{3}{2}.1}=-1\)
\(\left\{{}\begin{matrix}\left|x\right|=\left|y\right|\\x< 0\\y>0\end{matrix}\right.\) \(\Rightarrow y=-x\)
\(\Rightarrow\) \(x+y=x+\left(-x\right)=0\)
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}=\dfrac{0}{xy}=0\)
a) Ta có: \(A=1\dfrac{1}{4}\cdot x^3y\cdot\left(-\dfrac{6}{7}xy^5\right)^0\cdot\left(-2\dfrac{2}{3}xy\right)\)
\(=\dfrac{5}{4}x^3y\cdot\dfrac{-8}{3}xy\)
\(=\left(\dfrac{5}{4}\cdot\dfrac{-8}{3}\right)\cdot\left(x^3\cdot x\right)\cdot\left(y\cdot y\right)\)
\(=\dfrac{-10}{3}x^4y^2\)
ta có:
\(A=\dfrac{xy}{xy}-\dfrac{x-y}{y-x}.\left(\dfrac{x}{x}-\dfrac{y}{y}\right)\\ =1-\dfrac{x-y}{y-x}.\left(1-1\right)\\ =1-\dfrac{x-y}{y-x}.0\\ =1-0\\ =1\)
ta có:
xyxy−x−yy−x.(xx−yy)
\(1-\dfrac{x-y}{y-x}.\left(1-1\right)\\ =1-\dfrac{x-y}{y-x}.0\\=1-0\\ =1 \)