Question

[Kiều Thị Huyền]

 giải phương trình 

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

Answer 1

[๖Fly༉Donutღღ]

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

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 

Answer 2

[Không Tên]

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....