Question

\(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}-\frac{1}{x-1}\)

tìm ĐKXĐ

tìm x để giá trị trên bằng 0

 

Answer 1

\(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}-\frac{1}{x-1}\)

\(=\frac{\left(x-1\right)^2}{x^2-2x+1+3x}-\frac{1-2x^2+4x}{x^3-1}-\frac{1}{x-1}\)

\(=\frac{\left(x-1\right)^2}{x^2+x+1}-\frac{1-2x^2+4x}{x^3-1}-\frac{1}{x-1}\)

a)\(ĐKXĐ:x\ne1\)

\(MTC:\left(x-1\right)^3=\left(x-1\right)\left(x^2+x+1\right)\)

b)\(\frac{\left(x-1\right)^3}{x^3-1}-\frac{1-2x^2+4x}{x^3-1}-\frac{x^2+x+1}{x^3-1}=0\)

\(\Rightarrow\left(x-1\right)^3-\left(1-2x^2+4x\right)-\left(x^2+x+1\right)=0\)

\(\Leftrightarrow x^3-3x^2+27x-1-1+2x^2-4x-x^2-x-1=0\)

\(\Leftrightarrow x^3-2x^2+22x-3=0\)

ĐẾN ĐÂY THÌ BÍ RỒI T_T