a) 3(2^2+1)(2^4+1)(2^8+1)(2^16+1) b) x(2x^2-3) -x^2 (5x+1) + x^2 c) 3x(x-2) – 5x(1-x) – 8(x^2-3)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
3\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\\
 = \left( {{2^2} – 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\\
 = \left( {{2^4} – 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\\
 = \left( {{2^8} – 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\\
 = \left( {{2^{16}} – 1} \right)\left( {{2^{16}} + 1} \right)\\
 = {2^{32}} – 1\\
b,\\
x\left( {2{x^2} – 3} \right) – {x^2}\left( {5x + 1} \right) + {x^2}\\
 = 2{x^3} – 3x – 5{x^3} – {x^2} + {x^2}\\
 =  – 3{x^3} – 3x =  – 3x\left( {{x^2} + 1} \right)\\
c,\\
3x\left( {x – 2} \right) – 5x\left( {1 – x} \right) – 8\left( {{x^2} – 3} \right)\\
 = 3{x^2} – 6x – 5x + 5{x^2} – 8{x^2} + 24\\
 =  – 11x + 24
\end{array}\)