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ĐKXĐ : \(x,y\ne0\)\(;\)\(x\ne y\)
\(a)\) \(P=\frac{2}{x}-\left(\frac{x^2}{x^2-xy}+\frac{x^2-y^2}{xy}-\frac{y^2}{y^2-xy}\right):\frac{x^2-xy+y^2}{x-y}\)
\(P=\frac{2}{x}-\left(\frac{x^2y}{xy\left(x-y\right)}+\frac{\left(x-y\right)^2\left(x+y\right)}{xy\left(x-y\right)}+\frac{xy^2}{xy\left(x-y\right)}\right):\frac{x^2-xy+y^2}{x-y}\)
\(P=\frac{2}{x}-\left(\frac{xy\left(x+y\right)+\left(x-y\right)^2\left(x+y\right)}{xy\left(x-y\right)}\right):\frac{x^2-xy+y^2}{x-y}\)
\(P=\frac{2}{x}-\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{xy\left(x-y\right)}.\frac{x-y}{x^2-xy+y^2}\)
\(P=\frac{2y}{xy}-\frac{x+y}{xy}=\frac{y-x}{xy}\)
\(b)\)
+) Với \(\left|2x-1\right|=1\)\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-1=1\\2x-1=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}}\)
Mà \(x\ne0\) ( ĐKXĐ ) nên \(x=1\)
+) Với \(\left|y+1\right|=\frac{1}{2}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}y+1=\frac{1}{2}\\y+1=\frac{-1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}y=\frac{-1}{2}\\y=\frac{-3}{2}\end{cases}}}\)
Thay \(x=1;y=\frac{-1}{2}\) vào \(A=\frac{y-x}{xy}\) ta được : \(A=\frac{\frac{-1}{2}-1}{1.\frac{-1}{2}}=\frac{\frac{-3}{2}}{\frac{-1}{2}}=3\)
Thay \(x=1;y=\frac{-3}{2}\) vào \(A=\frac{y-x}{xy}\) ta được : \(A=\frac{\frac{-3}{2}-1}{1.\frac{-3}{2}}=\frac{\frac{-5}{2}}{\frac{-3}{2}}=\frac{15}{4}\)
Vậy ...
\(C=\frac{\left(x+y+2\right)^2}{xy+2\left(x+y\right)}+\frac{xy+2\left(x+y\right)}{\left(x+y+2\right)^2}=\frac{8}{9}.\frac{\left(x+y+2\right)^2}{xy+2\left(x+y\right)}+\frac{\left(x+y+2\right)^2}{9\left(xy+2x+2y\right)}+\frac{xy+2x+2y}{\left(x+y+2\right)^2}\)
\(C\ge\frac{4}{9}.\frac{2x^2+2y^2+4xy+8x+8x+8}{xy+2x+2y}+2\sqrt{\frac{\left(x+y+2\right)^2\left(xy+2x+2y\right)}{9\left(xy+2x+2y\right)\left(x+y+2\right)^2}}\)
\(C\ge\frac{4}{9}.\frac{\left(x^2+y^2\right)+\left(x^2+4\right)+\left(y^2+4\right)+4xy+8x+8y}{xy+2x+2y}+\frac{2}{3}\)
\(C\ge\frac{4}{9}.\frac{2xy+4x+4y+4xy+8x+8y}{xy+2x+2y}+\frac{2}{3}\)
\(C\ge\frac{4}{9}.\frac{6\left(xy+2x+2y\right)}{xy+2x+2y}+\frac{2}{3}=\frac{8}{3}+\frac{2}{3}=\frac{10}{3}\)
\(C_{min}=\frac{10}{3}\) khi \(x=y=2\)
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)`
`3x(2xy - 5x^2y)`
`= 3x*2xy + 3x* (-5x^2y)`
`= 6x^2y - 15x^3y`
`b)`
`2x^2y (xy - 4xy^2 + 7y)`
`= 2x^2y * xy + 2x^2y * (-4xy^2) + 2x^2y * 7y`
`= 2x^3y^2 - 8x^3y^3 + 14x^2y^2`
`c)`
`(-2/3xy^2 + 6yz^2)*(-1/2xy)`
`= (-2/3xy^2)*(-1/2xy) + 6yz^2 * (-1/2xy)`
`= 1/3x^2y^3 - 3xy^2z^2`
`a, 3x(2xy-5x^2y)`
`= 6x^2y - 15x^3y`
`b, 2x^2y(xy-4xy^2+7y)`
`= 2x^3y^2 - 8x^3y^3 + 14x^2y^2`
`c, (-2/3xy^2 + 6yz^2).(-1/2xy)`
`= 1/3x^2y^3 - 3xy^2z^2`
\(B=\left(\frac{x}{y^2+xy}-\frac{x-y}{x^2+xy}\right):\left(\frac{y^2}{x^3-xy^2}+\frac{1}{x+y}\right):\frac{3y}{y}\)
\(=\frac{x^2-xy+y^2}{xy\left(x+y\right)}\cdot\frac{x\left(x^2-y^2\right)}{x^2-xy+y^2}\cdot\frac{y}{3x}\)\(=\frac{x-y}{y}\cdot\frac{y}{3x}=\frac{x-y}{3x}\)