Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{\left( {x - 1} \right)^2} - \left( {x - 2} \right)\left( {x + 3} \right) + {\left( {x + 2} \right)^2} = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) + 6x.\left( {x + 2} \right)\\
\Leftrightarrow \left( {{x^2} - 2x + 1} \right) - \left( {{x^2} + x - 6} \right) + \left( {{x^2} + 4x + 4} \right) = \left( {{x^3} - {3^3}} \right) + 3{x^2} + 12x\\
\Leftrightarrow {x^2} - 2x + 1 - {x^2} - x + 6 + {x^2} + 4x + 4 = {x^3} - 27 + 6{x^2} + 12x\\
\Leftrightarrow {x^2} + x + 11 = {x^3} - 27 + 6{x^2} + 12x\\
\Leftrightarrow {x^3} + 5{x^2} + 11x - 38 = 0\\
b,\\
{\left( {x - 3} \right)^3} - \left( {2x + 1} \right)\left( {4{x^2} - 2x + 1} \right) = {\left( {x + 3} \right)^3} - {\left( {2x - 3} \right)^3} - 18x\left( {2x - 3} \right)\\
\Leftrightarrow \left( {{x^3} - 6{x^2} + 12x - 27} \right) - \left( {{{\left( {2x} \right)}^3} + {1^3}} \right) = \left( {{x^3} + 6{x^2} + 12x + 27} \right) - \left( {8{x^3} - 36{x^2} + 54x - 27} \right) - 36{x^2} + 54x\\
\Leftrightarrow - 6{x^2} - 27 - 8{x^3} - 1 = 6{x^2} + 27 - 8{x^3} + 27\\
\Leftrightarrow 12{x^2} + 82 = 0\,\,\,\,\,\,\,\left( {vn} \right)
\end{array}\)
Em xem lại đề câu a nhé!
