Đặt A =\(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^3-1}\)
ĐKXĐ: \(x\ne1\)
⇔ \(\frac{x^2+x+1+2x-2}{x^3-1}=\frac{3x^2}{x^3-1}\)
⇔ \(x^2+x+1+2x-2=3x^2\)
⇔ \(x^2+x+1+2x-2-3x^2=0\)
⇔ \(-2x^2+3x-1=0\)
⇔ \(-2x^2+2x+x-1=0\)
⇔ \(\left(x-1\right)\left(1-2x\right)=0\)
⇔ \(\left[{}\begin{matrix}x-1=0\\1-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\left(lo\text{ạ}i\right)\\x=\frac{1}{2}\left(tm\right)\end{matrix}\right.\) Vậy S = {\(\frac{1}{2}\)}
đkxđ \(x\ne1\)
\(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^3-1}\)
\(\Leftrightarrow\frac{1\left(x^2+x+1\right)+2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow\frac{x^2+x+1+x-1-3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{-2x^2+2x}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{-2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{-2x}{x^2+x+1}=0\)
\(\Rightarrow-2x=0\)
\(\Rightarrow x=0\left(t/m\right)\)
\(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^3-1}\)
\(\Leftrightarrow\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}\)
điều kiện xác định : x-1 ≠ 0 <=> x ≠ 1
Với x khác 1 ta có :
\(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Rightarrow x^2+x+1+2\left(x-1\right)=3x^2\)
\(\Leftrightarrow x^2+x+1+2x-2=3x^2\)
\(\Leftrightarrow x^2-3x^2+2x+x=2-1\)
\(\Leftrightarrow-2x^2+3x=1\)
\(\Leftrightarrow x\left(-2x+3\right)=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\-2x+3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)
\(\dfrac{1}{{x - 1}} + \dfrac{2}{{{x^2} + x + 1}} = \dfrac{{3{x^2}}}{{{x^3} - 1}}\)
ĐKXĐ: \(x \ne 1\)
\( \Leftrightarrow {x^2} + x + 1 + 2\left( {x - 1} \right) - 3{x^2} = 0\\ \Leftrightarrow {x^2} + x + 1 + 2x - 2 - 3{x^2} = 0\\ \Leftrightarrow - 2{x^2} + 3x - 1 = 0\\ \Leftrightarrow - 2{x^2} + 2x + x - 1 = 0\\ \Leftrightarrow - 2x\left( {x - 1} \right) + \left( {x - 1} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( { - 2x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 1\left( {ktm} \right)\\ x = \dfrac{1}{2}\left( {tm} \right) \end{array} \right. \)
