a) \(\dfrac{1}{{x - 1}} - \dfrac{{3{x^2}}}{{{x^3} - 1}} = \dfrac{{2x}}{{{x^2} + x + 1}}\)
Ta có: \({x^3} - 1 = \left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\)
\( = \left( {x - 1} \right)\left( {{x^2} + x + \dfrac{1}{4} + \dfrac{3}{4}} \right)\)
\( = \left( {x - 1} \right)\left[ {{x^2} + 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}} \right]\)
\( = \left( {x - 1} \right)\left[ {{{\left( {x + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}} \right]\)
Ta có: \({\left( {x + \dfrac{1}{2}} \right)^2} \geqslant 0\) với mọi \(x \in\mathbb R\) nên \({\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\) với mọi \(x \in\mathbb R\)
Do đó: \({x^3} - 1 \ne 0\) khi \(x - 1 ≠ 0⇔ x ≠ 1\)
ĐKXĐ: \(x ≠ 1\)
MTC: \({x^3} - 1\)
\( \Leftrightarrow \dfrac{{{x^2} + x + 1}}{{{x^3} - 1}} - \dfrac{{3{x^2}}}{{{x^3} - 1}} = \dfrac{{2x\left( {x - 1} \right)}}{{{x^3} - 1}}\)
\(\Rightarrow {x^2} + x + 1 - 3{x^2} = 2x\left( {x - 1} \right) \)
\(\Leftrightarrow - 2{x^2} + x + 1 = 2{x^2} - 2x\)
\( \Leftrightarrow 0 = 2{x^2} - 2x + 2{x^2} - x - 1\)
\( \Leftrightarrow 0 = 4{x^2} - 3x - 1\)
\(\Leftrightarrow 4{x^2} - 3x - 1 = 0\)
\(\Leftrightarrow 4{x^2} - 4x+x - 1 = 0\)
\(\Leftrightarrow 4x\left( {x - 1} \right) + \left( {x - 1} \right) = 0\)
\(\Leftrightarrow \left( {x - 1} \right)\left( {4x + 1} \right) = 0\)
\( \Leftrightarrow \left[ \begin{gathered} x - 1 = 0 \hfill \\ 4x + 1 = 0 \hfill \\ \end{gathered} \right.\)
\( \Leftrightarrow \left[ \begin{gathered} x = 1 \hfill \\ 4x = - 1 \hfill \\ \end{gathered} \right.\)
\(\Leftrightarrow \left[ {\matrix{{x = 1}\text{( loại)} \cr {x = - \dfrac{1}{4}}\text{(thỏa mãn)}\cr} }\right.\)
Vậy phương trình có nghiệm duy nhất \(x = - \dfrac{1}{4}\)
b) \(\dfrac{3}{{\left( {x - 1} \right)\left( {x - 2} \right)}} + \dfrac{2}{{\left( {x - 3} \right)\left( {x - 1} \right)}} \)\(\,= \dfrac{1}{{\left( {x - 2} \right)\left( {x - 3} \right)}}\)
ĐKXĐ: \(x ≠ 1, x ≠ 2, x ≠ 3\)
MTC: \((x-1)(x-2)(x-3)\)
\( \Rightarrow 3\left( {x - 3} \right) + 2\left( {x - 2} \right) = x - 1\)
\(\Leftrightarrow 3x - 9 + 2x - 4 = x - 1\)
\( \Leftrightarrow 5x - 13 = x - 1\)
\( \Leftrightarrow 5x - x = - 1 + 13\)
\(⇔ 4x = 12\)
\( \Leftrightarrow x = 12:4\)
\(⇔ x = 3\) (không thỏa mãn ĐKXĐ)
Vậy phương trình vô nghiệm.
c) \(1 + \dfrac{1}{{x + 2}} = \dfrac{{12}}{{8 + {x^3}}}\)
Ta có: \(8 + {x^3} ={2^3} + {x^3}\)\(\,= \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)\)
\( = \left( {x + 2} \right)\left( {{x^2} - 2x + 1 + 3} \right)\)
\( = \left( {x + 2} \right)\left[ {{{\left( {x - 1} \right)}^2} + 3} \right]\)
\({\left( {x - 1} \right)^2} \geqslant 0\) với mọi \(x\in \mathbb R\) nên \({\left( {x - 1} \right)^2} + 3 > 0\) với mọi \(x\in \mathbb R\)
Do đó: \(8 + {x^3} \ne 0\) khi \(x + 2 ≠ 0 ⇔ x ≠ -2\)
ĐKXĐ: \(x ≠ -2\)
MTC: \(8 + {x^3}\)
\( \Leftrightarrow \dfrac{{8 + {x^3}}}{{8 + {x^3}}} + \dfrac{{{x^2} - 2x + 4}}{{8 + {x^3}}} = \dfrac{{12}}{{8 + {x^3}}}\)
\( \Rightarrow {x^3} + 8 + {x^2} - 2x + 4 = 12 \)
\( \Leftrightarrow {x^3} + {x^2} - 2x = 12 - 8 - 4\)
\(\Leftrightarrow {x^3} + {x^2} - 2x = 0\)
\(\Leftrightarrow x\left( {{x^2} + x - 2} \right) = 0\)
\(\Leftrightarrow x\left[ {{x^2} + 2x - x - 2} \right] = 0\)
⇔\(x[ x(x+2) - (x+2) ] = 0\)
⇔ \(x(x + 2)(x - 1) = 0\)
\( \Leftrightarrow \left[ \begin{array}{l}x = 0\\x + 2 = 0\\x - 1 = 0\end{array} \right. \)
\(\Leftrightarrow \left[ \begin{array}{l}x = 0\left( \text{ thỏa mãn} \right)\\x = - 2\left( \text{ loại} \right)\\x = 1\left( \text{ thỏa mãn} \right)\end{array} \right.\)
Vậy phương trình có tập nghiệm là \(S = \left\{ {0;1} \right\}\).
d) \(\dfrac{{13}}{{\left( {x - 3} \right)\left( {2x + 7} \right)}} + \dfrac{1}{{2x + 7}} \)\(\,= \dfrac{6}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\)
ĐKXĐ: \(x \ne 3,x \ne - 3,x \ne - \dfrac{7}{2}\)
MTC: \({\left( {x - 3} \right)\left( {x + 3} \right)}\left( {2x + 7} \right)\)
\( \Rightarrow 13\left( {x + 3} \right) + \left( {x - 3} \right)\left( {x + 3} \right) \)\(= 6\left( {2x + 7} \right) \)
\(\Leftrightarrow 13x + 39 + {x^2} - 9 = 12x + 42\)
\(\Leftrightarrow {x^2} + 13x + 30 = 12x + 42\)
\( \Leftrightarrow {x^2} + 13x + 30 - 12x - 42 = 0\)
\(\Leftrightarrow {x^2} + x - 12 = 0\)
\(\Leftrightarrow {x^2} + 4x - 3x - 12 = 0\)
\(\Leftrightarrow x\left( {x + 4} \right) - 3\left( {x + 4} \right) = 0\)
\(\Leftrightarrow \left( {x - 3} \right)\left( {x + 4} \right) = 0\)
\( \Leftrightarrow \left[ \begin{array}{l}x - 3 = 0\\x + 4 = 0\end{array} \right. \)
\(\Leftrightarrow \left[ \begin{array}{l}x = 3\left( \text{không thỏa mãn} \right)\\x = - 4\left( \text{thỏa mãn} \right)\end{array} \right.\)
Vậy phương trình có nghiệm duy nhất \(x = -4\).
