Question

Giải phương trình sau:

2/x-1 - 3x^2/x^3-1 = x/x^2+x+1

Answer 1

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

ĐKXĐ : x khác 1

pt <=> \(\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

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

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

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

=> -2x² + 3x + 2 = 0

<=> -2x² - x + 4x + 2 = 0

<=> -x( 2x + 1 ) + 2( 2x + 1 ) = 0

<=> ( 2x + 1 )( 2 - x ) = 0

<=> x = -1/2 hoặc x = 2 ( tm )

Vậy ...

Answer 2

\(\frac{2}{x-1}-\frac{3x^2}{x^3-1}=\frac{x}{x^2+x+1}\)ĐK : x \(\ne\)1

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

\(\Rightarrow2x^2+2x+2-3x^2=x^2-x\)

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

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

Vậy tập nghiệm của phương trình là S = { 1/2 ; 2 }