Question

Giải phương trình sau

Answer 1

PT \(\Leftrightarrow\dfrac{x^2+x+1-3x^2-2x^2+2x}{\left(x-1\right)\left(x^2+x+1\right)}=0\left(x\ne1\right)\)

 \(\Leftrightarrow-4x^2+3x+1=0\)

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

\(\Leftrightarrow-4x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=-\dfrac{1}{4}\left(TMx\ne1\right)\end{matrix}\right.\)

Answer 2

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

ĐKXĐ: x≠1

<=> \(\dfrac{x^2+x+1}{x^3-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x^2-2x}{x^3-1}\)

=> x²+x+1-3x²=2x²-2x

<=> x²-2x²+x+2x-3x²=-1 

<=> -4x²+3x-1=0 <=> 4x²-3x+1=0 

<=> 4x²-2.2x.\(\dfrac{3}{4}\)+\(\dfrac{9}{16}+\dfrac{7}{16}\)=0

<=> (2x-\(\dfrac{3}{4}\))²=-\(\dfrac{7}{16}\) (KTMĐK)

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