Câu hỏi của Quốc An

Question

e ) $\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}$

Trần Việt Linh · Accepted Answer

e) $\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}$$=\frac{x}{x-2y}+\frac{x}{x+2y}-\frac{4xy}{\left(x-2y\right)\left(x+2y\right)}$$=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}$$=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}$$=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}$$=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x}{x+2y}$Câu trả lời: e) $\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}$$=\frac{x}{x-2y}+\frac{x}{x+2y}-\frac{4xy}{\left(x-2y\right)\left(x+2y\right)}$$=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}$$=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}$$=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}$$=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x}{x+2y}$

Võ Đông Anh Tuấn · Answer

e ) $\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}$$=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}$$=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}$$=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}$$=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}$$=\frac{2x}{x+2y}$Câu trả lời: e ) $\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}$$=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}$$=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}$$=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}$$=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}$$=\frac{2x}{x+2y}$

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