Câu hỏi của Lê Huỳnh Thanh Ngân

Question

$\frac{\left(x+y\right)^2}{x}.\left[\frac{x}{\left(x+y\right)^2}-\frac{x}{x^2-y^2}\right]-\frac{5x-3y}{y-x}$Chứng minh biểu thức sau ko phụ thuộc vào biến

Nguyễn Nam Dương · Accepted Answer

$\frac{\left(x+y\right)^2}{x}.\left(\frac{x}{\left(x+y\right)^2}-\frac{x}{x^2-y^2}\right)-\frac{5x-3y}{y-x}\left(đk:x\text{≠}0-y;y\right).$$=\frac{\left(x+y\right)^2}{x}.\left(\frac{x}{\left(x+y\right)^2}-\frac{x}{\left(x-y\right)\left(x+y\right)}\right)-\frac{5x-3y}{y-x}$$=\frac{\left(x+y\right)^2}{x}.\frac{x\left(x-y\right)-x\left(x+y\right)}{\left(x+y\right)^2\left(x-y\right)}+\frac{5x-3y}{x-y}$$=\frac{1}{x}.\frac{x^2-xy-x^2-xy}{\left(x+y\right)^2\left(x-y\right)}+\frac{5x-3y}{x-y}$$=\frac{1}{x}.\frac{-2xy}{x-y}+\frac{5x-3y}{x-y}$$=\frac{-2y}{x-y}+\frac{5x-3y}{x-y}$$=\frac{-2xy+5x-3y}{x-y}$$=\frac{5\left(x-y\right)}{x-y}$$=5$Ta có đpcmCâu trả lời: $\frac{\left(x+y\right)^2}{x}.\left(\frac{x}{\left(x+y\right)^2}-\frac{x}{x^2-y^2}\right)-\frac{5x-3y}{y-x}\left(đk:x\text{≠}0-y;y\right).$$=\frac{\left(x+y\right)^2}{x}.\left(\frac{x}{\left(x+y\right)^2}-\frac{x}{\left(x-y\right)\left(x+y\right)}\right)-\frac{5x-3y}{y-x}$$=\frac{\left(x+y\right)^2}{x}.\frac{x\left(x-y\right)-x\left(x+y\right)}{\left(x+y\right)^2\left(x-y\right)}+\frac{5x-3y}{x-y}$$=\frac{1}{x}.\frac{x^2-xy-x^2-xy}{\left(x+y\right)^2\left(x-y\right)}+\frac{5x-3y}{x-y}$$=\frac{1}{x}.\frac{-2xy}{x-y}+\frac{5x-3y}{x-y}$$=\frac{-2y}{x-y}+\frac{5x-3y}{x-y}$$=\frac{-2xy+5x-3y}{x-y}$$=\frac{5\left(x-y\right)}{x-y}$$=5$Ta có đpcm

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