rút gon biểu thức sau A=(2x+1)^2+2(4x^2-1)(2x-1)^1 làm chi tiết giúp em thank ạ 24/09/2021 Bởi Madeline rút gon biểu thức sau A=(2x+1)^2+2(4x^2-1)(2x-1)^1 làm chi tiết giúp em thank ạ
\[\begin{array}{l} A = {\left( {2x + 1} \right)^2} + 2\left( {4{x^2} – 1} \right)\left( {2x – 1} \right)\\ = {\left( {2x + 1} \right)^2} + 2\left( {2x – 1} \right)\left( {2x + 1} \right)\left( {2x – 1} \right)\\ = {\left( {2x + 1} \right)^2} + 2\left( {2x + 1} \right){\left( {2x – 1} \right)^2}\\ = \left( {2x + 1} \right)\left[ {2x + 1 + 2\left( {4{x^2} – 4x + 1} \right)} \right]\\ = \left( {2x + 1} \right)\left( {2x + 1 + 8{x^2} – 8x + 2} \right)\\ = \left( {2x + 1} \right)\left( {8{x^2} – 6x + 3} \right). \end{array}\] Bình luận
\[\begin{array}{l}
A = {\left( {2x + 1} \right)^2} + 2\left( {4{x^2} – 1} \right)\left( {2x – 1} \right)\\
= {\left( {2x + 1} \right)^2} + 2\left( {2x – 1} \right)\left( {2x + 1} \right)\left( {2x – 1} \right)\\
= {\left( {2x + 1} \right)^2} + 2\left( {2x + 1} \right){\left( {2x – 1} \right)^2}\\
= \left( {2x + 1} \right)\left[ {2x + 1 + 2\left( {4{x^2} – 4x + 1} \right)} \right]\\
= \left( {2x + 1} \right)\left( {2x + 1 + 8{x^2} – 8x + 2} \right)\\
= \left( {2x + 1} \right)\left( {8{x^2} – 6x + 3} \right).
\end{array}\]