Rút gọn: $\left(\dfrac{1}{3}+\dfrac{3}{x^2-3x}\right)\div \left(\dfrac{x^2}{27-3x^2}+\dfrac{1}{x+3}\right)$ 17/07/2021 Bởi Katherine Rút gọn: $\left(\dfrac{1}{3}+\dfrac{3}{x^2-3x}\right)\div \left(\dfrac{x^2}{27-3x^2}+\dfrac{1}{x+3}\right)$
Đặt `A=(1/3+3/(x^2-3x)):(x^2/(27-3x^2)+1/(x+3))` `đk:x ne 0,x ne -3` `A=(1/3+3/(x(x-3))):(x^2/(3(3-x)(3+x))+1/(x+3))` `=((x^2-3x+9)/(3x(x-3))):((3(x^2-9)-x^2)/(3(x-3)(x+3)))` `=((x^2-3x+9)/(3x(x-3))).(3(x-3)(x+3))/(2x^2-27)` `=((x^2-3x+9)(x+3))/(x(2x^2-27))` Bình luận
Đặt `A=(1/3+3/(x^2-3x)):(x^2/(27-3x^2)+1/(x+3))`
`đk:x ne 0,x ne -3`
`A=(1/3+3/(x(x-3))):(x^2/(3(3-x)(3+x))+1/(x+3))`
`=((x^2-3x+9)/(3x(x-3))):((3(x^2-9)-x^2)/(3(x-3)(x+3)))`
`=((x^2-3x+9)/(3x(x-3))).(3(x-3)(x+3))/(2x^2-27)`
`=((x^2-3x+9)(x+3))/(x(2x^2-27))`
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