Question

tính giá trị của biểu thức
\(\frac{2}{y-x}\) - \(\left(\frac{2xy}{x^2-y^2}+\frac{x-y}{2x+2y}\right)\): \(\frac{x^2+xy}{2x}\) tại x-y = 2011
giúp mình với mn ơi ✿

Answer 1

Giả sử biểu thức xác định

\(\frac{-2}{x-y}-\left(\frac{2xy}{\left(x-y\right)\left(x+y\right)}+\frac{x-y}{2\left(x+y\right)}\right).\frac{2x}{x\left(x+y\right)}\)

\(=\frac{-2}{x-y}-\left(\frac{4xy+\left(x-y\right)^2}{2\left(x-y\right)\left(x+y\right)}\right).\frac{2}{x+y}\)

\(=\frac{-2}{x-y}-\frac{\left(x+y\right)^2}{2\left(x-y\right)\left(x+y\right)}.\frac{2}{x+y}\)

\(=\frac{-2}{x-y}-\frac{1}{x-y}=\frac{-3}{x-y}=\frac{-3}{2011}\)

Answer 2

\(\frac{2}{y-x}\cdot\left(\frac{2xy}{x^2-y^2}+\frac{x-y}{2x+2y}\right):\frac{x^2+xy}{2x}=\frac{2}{y-x}\cdot\left(\frac{4xy}{2\left(x-y\right)\left(x+y\right)}+\frac{\left(x-y\right)^2}{2\left(x+y\right)\left(x-y\right)}\right):\frac{x^2+xy}{2x}\)

\(=\frac{2}{y-x}\cdot\left(\frac{4xy+x^2-2xy+y^2}{2\left(x+y\right)\left(x-y\right)}\right)\cdot\frac{2x}{x\left(x+y\right)}=\frac{2}{y-x}\cdot\frac{\left(x+y\right)^2\cdot2x}{2\left(x+y\right)\left(x-y\right)\cdot x\left(x+y\right)}=\frac{2}{y-x}\cdot\frac{1}{x-y}\)

\(=\frac{2}{-\left(x-y\right)}\cdot\frac{1}{x-y}\)

Mà x - y = 2011

\(\Rightarrow\frac{2}{-\left(x-y\right)}+\frac{1}{x-y}=\frac{-2}{2011}+\frac{1}{2011}=\frac{-1}{2011}\)