Gọi \(x_1\) là nghiệm của phương trình \(x^3 + 2(x - 1) ^2 - 2(x - 1)(x + 1) = x^3 + x - 4 - (x - 4) \). Tính \(x_1.x_2\)

* Hướng dẫn giải

+ Ta có

\(\begin{array}{l} {x^3} + 2{\left( {x - 1} \right)^2} - 2\left( {x - 1} \right)\left( {x + 1} \right) = {x^3} + x - 4 - \left( {x - 4} \right)\\ \Leftrightarrow {x^3} + 2{\left( {x - 1} \right)^2} - 2\left( {x - 1} \right)\left( {x + 1} \right) - {x^3} - x + 4 + \left( {x - 4} \right) = 0\\ \Leftrightarrow \left( {{x^3} - {x^3}} \right) + 2\left( {{x^2} - 2x + 1} \right) - 2\left( {{x^2} - 1} \right) - x + 4 + x - 4 = 0\\ \Leftrightarrow 2{x^2} - 4x + 2 - 2{x^2} + 2 - x + 4 + x - 4 = 0\\ \Leftrightarrow \left( {2{x^2} - 2{x^2}} \right) + \left( { - 4x - x + x} \right) + \left( {2 + 2 + 4 - 4} \right) = 0\\ \begin{array}{*{20}{l}} { \Leftrightarrow - 4x + 4 = 0}\\ { \Leftrightarrow - 4x = - 4}\\ { \Leftrightarrow x = 1} \end{array} \end{array}\)

Suy ra \(x_1=1\)

+ Ta có \(\begin{array}{l} x + \frac{{2x - 7}}{2} = 5 - \frac{{x + 6}}{2} + \frac{{3x + 1}}{5}\\ \Leftrightarrow \frac{{10x}}{{10}} + \frac{{5\left( {2x - 7} \right)}}{{10}} = \frac{{50}}{{10}} - \frac{{5\left( {x + 6} \right)}}{{10}} + \frac{{2\left( {3x + 1} \right)}}{{10}}\\ \begin{array}{*{20}{l}} { \Leftrightarrow 20x - 35 = x + 22}\\ { \Leftrightarrow 20x - x = 22 + 35}\\ { \Leftrightarrow 19x = 57}\\ { \Leftrightarrow x = 57:19}\\ { \Leftrightarrow x = 3} \end{array} \end{array}\)

Nên \(x_1.x_2=1.3=3\)