Đáp án:
Giải thích các bước giải:
a) `E=\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}`
ĐK: `x \ne +- 2y`
`E=\frac{x(x+2y)}{(x-2y)(x+2y)}+\frac{x(x-2y)}{(x-2y)(x+2y)}-\frac{4xy}{(x-2y)(x+2y)}`
`E=\frac{x^2+2xy}{(x-2y)(x+2y)}+\frac{x^2-2xy}{(x-2y)(x+2y)}-\frac{4xy}{(x-2y)(x+2y)}`
`E=\frac{x^2+2xy+x^2-2xy-4xy}{(x-2y)(x+2y)}`
`E=\frac{2x^2-4xy}{(x-2y)(x+2y)}`
`E=\frac{2x(x-2y)}{(x-2y)(x+2y)}`
`E=\frac{2x}{x+2y}`
b) `F=(\frac{x^4+x^2-4x+1}{x^2-1}-\frac{x-1}{x+1}+\frac{x+1}{x-1}).\frac{x(x+1)-1-x}{x^3-1}`
ĐK: `x \ne +- 1`
`F=[\frac{x^4+x^2-4x+1}{(x-1)(x+1)}-\frac{(x-1)^2}{(x-1)(x+1)}+\frac{(x+1)^2}{(x-1)(x+1)}].\frac{x(x+1)-(1+x)}{x^3-1}`
`F=[\frac{x^4+x^2-4x+1}{(x-1)(x+1)}-\frac{x^2-2x+1}{(x-1)(x+1)}+\frac{x^2+2x+1}{(x-1)(x+1)}].\frac{(x-1)(x+1)}{(x-1)(x^2+x+1)}`
`F=[\frac{x^4+x^2-4x+1-x^2+2x-1+x^2+2x+1}{(x-1)(x+1)}].\frac{(x-1)(x+1)}{(x-1)(x^2+x+1)}`
`F=\frac{x^4+x^2+1}{(x-1)(x+1)}.\frac{(x-1)(x+1)}{(x-1)(x^2+x+1)}`
`F=\frac{(x^2-x+1)(x^2+x+1)}{(x-1)(x+1)}.\frac{(x-1)(x+1)}{(x-1)(x^2+x+1)}`
`F=\frac{x^2-x+1}{x-1}`
