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a) Rút gọn :
Ta có : \(A=\frac{y-x}{xy}:\left[\frac{y^2}{\left(x-y\right)^2}-\frac{2x^2y}{\left(x^2-y^2\right)^2}+\frac{x^2}{y^2-x^2}\right]\)
\(=\frac{y-x}{xy}:\left[\frac{y^2\left(x+y\right)^2-2x^2y-x^2\left(x^2-y^2\right)}{\left(x^2-y^2\right)^2}\right]\)
\(=\frac{y-x}{xy}:\left[\frac{y^2\left(x^2+2xy+y^2\right)-2x^2y-x^4+x^2y^2}{\left(x^2-y^2\right)^2}\right]\)
...
Với đk trên ta có:
P = \(\frac{2}{x}-\left(\frac{x^2}{x^2+xy}+\frac{y^2-x^2}{xy}-\frac{y^2}{xy+y^2}\right).\frac{x+y}{x^2+xy+y^2}\)
\(=\frac{2}{x}-\left(\frac{x}{x+y}-\frac{\left(x-y\right)\left(x+y\right)}{xy}-\frac{y}{x+y}\right).\frac{x+y}{x^2+xy+y^2}\)
\(=\frac{2}{x}-\left(\frac{x-y}{x+y}-\frac{\left(x-y\right)\left(x+y\right)}{xy}\right).\frac{x+y}{x^2+xy+y^2}\)
\(=\frac{2}{x}-\frac{x-y}{xy}.\left(xy-\left(x+y\right)^2\right).\frac{1}{x^2+xy+y^2}\)
\(=\frac{2}{x}+\frac{x-y}{xy}\)
\(=\frac{x+y}{xy}\)
Ta có:
\(P=\frac{x^2-2xy+y^2}{x^2-2xy+y^2}\)
\(\Leftrightarrow P=\frac{\left(x^2+y^2\right)-2xy}{\left(x^2+y^2\right)+2xy}\) (1)
Mà \(\frac{xy}{x^2+y^2}=\frac58\)
\(\Rightarrow x^2+y^2=\frac58xy\) (2)
Thay (1) và (2) ta có:
\(P=\frac{\frac58xy-2xy}{\frac58xy+2xy}\)
\(\Rightarrow P=\frac{-\frac{11}{8}xy}{\frac{21}{8}xy}\)
\(\Rightarrow P=-\frac{11}{21}\)
Ta có:
\(P=\dfrac{xy}{x^2+y^2}=\dfrac58\Rightarrow x^2+y^2=\dfrac{8xy}{5}\)
\(\Rightarrow P=\dfrac{x^2-2xy+y^2}{x^2+2xy+y^2}\Leftrightarrow P=\dfrac{\left(x^2+y^2\right)-2xy}{\left(x^2+y_{}^2\right)+2xy}\)
Thay \(x^2+y^2=\dfrac{8xy}{5}\) vào \(P\) , có:
\(P=\dfrac{\left(x^2+y^2\right)-2xy}{\left(x^2+y^2\right)+2xy}\)
\(\Leftrightarrow P=\dfrac{\dfrac{8xy}{5}-2xy}{\dfrac{8xy}{5}+2xy}\)
\(\Leftrightarrow P=\dfrac{\dfrac{-2xy}{5}}{\dfrac{18xy}{5}}\)
\(\Leftrightarrow P=\dfrac{-2xy}{18xy}\)
\(\Leftrightarrow P=-\dfrac19\)
Vậy \(P=-\dfrac19\)
Đặt \(t = \frac{x}{y}\)
\(\frac{t}{t^{2} + 1} = \frac{5}{8}\)
\(8 t = 5 \left(\right. t^{2} + 1 \left.\right)\)
\(t + \frac{1}{t} = \frac{8}{5}\)
\(P = \frac{x^{2} - 2 x y + y^{2}}{x^{2} + 2 x y + y^{2}} = \frac{\left(\right. t - 1 \left.\right)^{2}}{\left(\right. t + 1 \left.\right)^{2}}\)
\(\left(\right. t - 1 \left.\right)^{2} = t \left(\right. \frac{8}{5} - 2 \left.\right) = - \frac{2}{5} t\)
\(\left(\right. t + 1 \left.\right)^{2} = t \left(\right. \frac{8}{5} + 2 \left.\right) = \frac{18}{5} t\)
\(P = \frac{- \frac{2}{5} t}{\frac{18}{5} t} = - \frac{1}{9}\)