Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
1.ĐK:x \ne \pm 1\\
\frac{{{x^2} + 2x + 1 + 2 - {x^2} + x}}{{(x - 1)(x + 1)}} = 0 \to \frac{{3x + 3}}{{(x - 1)(x + 1)}} = 0\\
\to \frac{3}{{x - 1}} = 0(vôlí)
\end{array}\)
⇒ Pt vô nghiệm
2. \(\begin{array}{l}
ĐK:x \ne 1\\
\frac{{2{x^3} - 2 + x - \left( {{x^2} + 1} \right)\left( {x - 1} \right) - x\left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = 0\\
\to \frac{{2{x^3} - 2 + x - {x^3} + {x^2} - x + 1 - {x^3} - {x^2} - x}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = 0\\
\to \frac{{ - x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = 0\\
\to x = - 1(TM)
\end{array}\)
3. \(\begin{array}{l}
ĐK:x \ne \pm 1\\
Pt \to \frac{{{x^4} + {x^2} - {x^2} + 1 - 2}}{{{x^4} - 1}} = 0\\
\to \frac{{{x^4} - 1}}{{{x^4} - 1}} = 0 \to 1 = 0(vôlí)
\end{array}\)
⇒ Pt vô nghiệm
4. \(\begin{array}{l}
ĐK:x \ne 3;x \ne 2;x \ne 5\\
Pt \to \frac{x}{{(x - 3)\left( {x - 2} \right)}} + \frac{x}{{(x - 5)(x - 2)}} - \frac{1}{{(x - 5)(x - 3)}} = 0\\
\to \frac{{{x^2} - 5x + {x^2} - 3x - x + 2}}{{(x - 3)\left( {x - 2} \right)\left( {x - 5} \right)}} = 0\\
\to 2{x^2} - 9x + 2 = 0 \to x = \frac{{9 \pm \sqrt {65} }}{4}(TM)
\end{array}\)
5. \(\begin{array}{l}
Pt \to (x - 2)(3x - 1) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = \frac{1}{3}
\end{array} \right.
\end{array}\)
6. \(\begin{array}{l}
ĐK:x \ne 1;x \ne - 4\\
Pt \to \frac{{15x - 12x + 12 - 4x - 16 - {x^2} - 3x + 4}}{{(x + 4)(x - 1)}} = 0\\
\to \frac{{x(x + 4)}}{{(x + 4)(x - 1)}} = 0 \to x = 0(TM)
\end{array}\)
