câu hỏi : a)(x-1)^2-(x-2)(x+3)+(x+2)^2=(x-3)(x^2+3x+9)+6x(x+2) b)(x-3)^3-(2x+1)(4x^2-2x+1)=(x+3)^3-(2x-3)^3-18x(2x-3)

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!