Đáp án:
Giải thích các bước giải:
a) `x/(x^3+1);(x+1)/(x^2+x);(x+2)/(x^2-x+1)`
`ĐK: x \ne -1`
`MTC: x(x+1)(x^2-x+1)`
`x/(x^3+1)=\frac{x^2}{x(x+1)(x^2-x+1)}`
`\frac{x+1}{x^2+x}=\frac{(x+1)(x^2-x+1)}{x(x+1)(x^2-x+1)}`
`\frac{x+2}{x^2-x+1}=\frac{x(x+1)(x+2)}{x(x+1)(x^2-x+1)}`
b) `\frac{x-1}{x+1};\frac{x+1}{x-1};\frac{1}{x^2-1}`
`ĐK: x \ne ±1`
`MTC: (x+1)(x-1)`
`\frac{x-1}{x+1}=\frac{(x-1)^2}{(x+1)(x-1)}`
`\frac{x+1}{x-1}=\frac{(x+1)^2}{(x+1)(x-1)}`
c) `\frac{x}{x^3-xy^2};\frac{1}{(x+y)^2};\frac{1}{(x-y)^2}`
`ĐK: x \ne ±y`
`MTC: x(x+y)^2. (x-y)^2`
`\frac{x}{x^3-xy^2}=\frac{x(x-y)(x+y)}{x(x+y)^2. (x-y)^2}`
`\frac{1}{(x+y)^2}=\frac{x(x-y)^2}{x(x+y)^2. (x-y)^2}`
`\frac{1}{(x-y)^2}=\frac{x(x+y)^2}{x(x+y)^2. (x-y)^2}`
