nhóm x ra ở phân thức 2 rồ quy đồng đi rồi cộng
\(\dfrac{x}{x-3}+\dfrac{9-6x}{x^4-3x}\)
\(=\dfrac{x\left(x^4-3x\right)}{\left(x-3\right)\left(x^4-3x\right)}+\dfrac{\left(9-6x\right)\left(x-3\right)}{\left(x-3\right)\left(x^4-3x\right)}\)
\(=\dfrac{x\left(x^4-3x\right)+\left(9-6x\right)\left(x-3\right)}{\left(x-3\right)\left(x^4-3x\right)}\)
\(=\dfrac{x^5-3x^2+9x-27-6x^2+18}{\left(x-3\right)\left(x^4-3x\right)}\)
\(=\dfrac{x^5-9x^2+9x-9}{\left(x-3\right)\left(x^4-3x\right)}\)
Sửa đề:
\(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}\)
\(=\dfrac{x}{x-3}+\dfrac{9-6x}{x\left(x-3\right)}\)
\(=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}\)
\(=\dfrac{x^2+9-6x}{x\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}\)
\(=\dfrac{x-3}{x}\)
\(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}\)
\(=\dfrac{x.x}{(x-3).x}+\dfrac{9-6x}{x\left(x-3\right)}\)
\(=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)