Câu hỏi của Khánh Linh

Question

rút gọn bt {[1/(x^2+2xy+y^2)]-[1/(x^2-y^2)]}/(4xy/y^2-x^2)

Thu Thao · Accepted Answer

$\dfrac{1}{x^2+2xy+y^2}-\dfrac{1}{x^2-y^2}:\dfrac{4xy}{y^2-x^2}$ $\left(x,y\ne0;x\ne\pm y\right)$$=\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{y^2-x^2}.\dfrac{y^2-x^2}{4xy}$$=\dfrac{1}{x^2+2xy+y^2}+\dfrac{1}{4xy}$$=\dfrac{6xy+x^2+y^2}{4xy\left(x+y\right)^2}$Câu trả lời: $\dfrac{1}{x^2+2xy+y^2}-\dfrac{1}{x^2-y^2}:\dfrac{4xy}{y^2-x^2}$ $\left(x,y\ne0;x\ne\pm y\right)$$=\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{y^2-x^2}.\dfrac{y^2-x^2}{4xy}$$=\dfrac{1}{x^2+2xy+y^2}+\dfrac{1}{4xy}$$=\dfrac{6xy+x^2+y^2}{4xy\left(x+y\right)^2}$

Nguyễn Lê Phước Thịnh · Answer

Ta có: $\dfrac{1}{x^2+2xy+y^2}-\dfrac{1}{x^2-y^2}:\dfrac{4xy}{y^2-x^2}$$=\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x+y\right)\left(x-y\right)}{4xy}$$=\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{4xy}$$=\dfrac{4xy}{4xy\left(x+y\right)^2}+\dfrac{x^2+2xy+y^2}{4xy\left(x+y\right)^2}$$=\dfrac{x^2+6xy+y^2}{4xy\left(x+y\right)^2}$Câu trả lời: Ta có: $\dfrac{1}{x^2+2xy+y^2}-\dfrac{1}{x^2-y^2}:\dfrac{4xy}{y^2-x^2}$$=\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x+y\right)\left(x-y\right)}{4xy}$$=\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{4xy}$$=\dfrac{4xy}{4xy\left(x+y\right)^2}+\dfrac{x^2+2xy+y^2}{4xy\left(x+y\right)^2}$$=\dfrac{x^2+6xy+y^2}{4xy\left(x+y\right)^2}$

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