Đáp án:
$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)=2^{32}-1$
$x(2x^2-3)-x^2(5x+1)+x^2=x(-3x^2-4x-3)$
$3x(x-2)-5x(1-x)-8(x^2-3)=-11x+24$
Giải thích các bước giải:
$\begin{split}A&=3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)\\&=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)\\&=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)\\&=(2^8-1)(2^8+1)(2^{16}+1)\\&=(2^{16}-1)(2^{16}+1)\\&=2^{32}-1\end{split}$
$\begin{split}x(2x^2-3)-x^2(5x+1)+x^2&=x(2x^2-3-x(5x+1)+x)\\&=x(2x^2-3-5x^2-5x+x)\\&=x(-3x^2-4x-3)\end{split}$
$\begin{split}3x(x-2)-5x(1-x)-8(x^2-3)&=3x^2-6x-5x+5x^2-8x^2+24\\&=-11x+24\end{split}$
