Ta có : \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
\(x^2+x=x\left(x+1\right)\)
\(x^2+x+1=x^2+x+1\)
MTC : \(x\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)
Quy đồng :
\(\frac{x}{x^3-1}=\frac{x}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)}\)
\(\frac{x+1}{x^2+x}=\frac{x+1}{x\left(x+1\right)}=\frac{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}{x\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{x-1}{x^2+x+1}=\frac{\left(x-1\right)^2\left(x+1\right)x}{x\left(x+1\right)\left(x^2+x+1\right)\left(x-1\right)}\)
\(\frac{x}{x^3-1};\frac{x+1}{x^2+x};\frac{x-1}{x^2+x+1}\)
Ta có:\(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
\(x^2+x=x\left(x+1\right)\)
\(x^2+x+1=x^2+x+1\)
\(\Rightarrow MTC=x\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)
Quy đồng:
\(\frac{x}{x^3-1}=\frac{x}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2\left(x+1\right)}{x\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)}\)
\(\frac{x+1}{x^2+x}=\frac{x+1}{x\left(x+1\right)}=\frac{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}{x\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{x-1}{x^2+x+1}=\frac{\left(x-1\right)^2x\left(x+1\right)}{x\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}\)
