\(\frac{2}{x-1}+\frac{2x+3}{x^2+x+1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)
Quy đồng mẫu chung :
\(\frac{2.\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{\left(4x^2-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
Sau đó ta khử mẫu:
\(\Rightarrow\)\(2x^2+2x+2+2x^2+x-3=4x^2-1\)
\(\Rightarrow\)\(2x^2+2x+2x^2+x-4x^2=-1-2+3\)
\(\Rightarrow\)\(3x=0\)
\(\Rightarrow\)\(x=0\)
Vậy bạn tự kết luận
ĐKXĐ: \(x\ne1\)
\(\frac{2}{x-1}+\frac{2x+3}{x^2+x+1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)
\(\Leftrightarrow\)\(\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{4x^2-1}{x^3-1}\)
\(\Leftrightarrow\)\(\frac{2x^2+2x+2}{x^3-1}+\frac{2x^2+x-3}{x^3-1}=\frac{4x^2-1}{x^3-1}\)
\(\Rightarrow\)\(2x^2+2x+2+2x^2+x-3=4x^2-1\)
\(\Leftrightarrow\)\(4x^2+3x-1=4x^2-1\)
\(\Leftrightarrow\)\(3x=0\)
\(\Leftrightarrow\)\(x=0\) (thỏa mãn)
Vậy....
\(\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x+3\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\)\(\frac{4x^2-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow2x^2+2x+2+2x^2-2x+3x-3\)\(=4x^2-1\)
\(\Leftrightarrow2x^2+2x^2-4x^2+2x-2x+3x=-1-2+3\)
\(\Leftrightarrow3x=0\)
\(\Leftrightarrow x=0\left(nhận\right)\)
Vậy S= {0}
