bạn ơi!! 2⁸ +1 chứ, bn xem lại đề cấy

\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

3.(2²+1).(2⁴+1).(2⁸+1).(2¹⁶+1)

=(2²-1).(2²+1).(2⁴+1).(2⁸+1).(2¹⁶+1)

=(2⁴-1).(2⁴+1).(2⁸+1).(2¹⁶+1)

=(2⁸-1).(2⁸+1).(2¹⁶+1)

=(2¹⁶-1).(2¹⁶+1)

=2³²-1

4294967295

cái này là rút gọn biểu thức

( bài này áp dụng hằng đẳng thức \(a^2-b^2=\left(a+b\right)\left(a-b\right)\)

Ta có

\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)

\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\)

\(=2^{64}-1\)

3.(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1)(2³²+1)

=(2²-1)(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1)(2³²+1)

=(2⁴-1)(2⁴+1)(2⁸+1)(2¹⁶+1)(2³²+1)

=(2⁸-1)(2⁸+1)(2¹⁶+1)(2³²+1)

=(2¹⁶-1)(2¹⁶+1)(2³²+1)

=(2³²-1)(2³²+1)

=2⁶⁴-1

bài này nâng cao 6 có đấy

   3(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1)(2³²+1)(2⁶⁴+1)

=(2²-1)(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1)(2³²+1)(2⁶⁴+1)

=(2⁴-1)(2⁴+1)(2⁸+1)(2¹⁶+1)(2³²+1)(2⁶⁴+1)

=(2⁸-1)(2⁸+1)(2¹⁶+1)(2³²+1)(2⁶⁴+1)

=(2¹⁶-1)(2¹⁶+1)(2³²+1)(2⁶⁴+1)

=(2³²-1)(2³²+1)(2⁶⁴+1)

=(2⁶⁴-1)(2⁶⁴+1)

=(2¹²⁸-1)

Nếu thấy đúng thì thích cho mình nha

\(B=\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\frac{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\)

\(=\frac{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\)

\(=\frac{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\)

\(=\frac{\left(2^{16}-1\right)\left(2^{16}+1\right)}{3}\)

\(=\frac{2^{32}-1}{3}\)

3(2² + 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1)

=(4 - 1)(2² + 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1)

=(2² - 1)(2² + 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1)

=(2⁴ - 1)(2⁴ + 1)(2⁸ + 1)(2¹⁶ + 1)

=(2⁸ - 1)(2⁸ + 1)(2¹⁶ + 1)

=(2¹⁶ - 1)(2¹⁶ + 1)

=(2³² - 1)

mình áp dụng hằng đẳng thức A² - B²=(A - B)(A + B)

Đặt A=3(2² +1)(2⁴+1)(2⁸+1)(2¹⁶+1)

=(4-1)(2²+1)(2⁴+1)(2⁸+1)(2¹⁶+1)

=[(2²-1)(2²+1)](2⁴+1)(2⁸+1)(2¹⁶+1)

=(2⁴-1)(2⁴+1)(2⁸+1)(2¹⁶+1)

=(2⁸-1)(2⁸+1)(2¹⁶+1)

=(2¹⁶-1)(2¹⁶+1)

=2³²-1

3(2² +1)(2⁴+1)(2⁸+1)(2¹⁶+1) = (2² -1)(2² +1)(2⁴+1)(2⁸+1)(2¹⁶+1) = (2⁴-1)(2⁴+1)(2⁸+1)(2¹⁶+1) = (2⁸-1)(2⁸+1)(2¹⁶+1) 

= (2¹⁶-1)(2¹⁶+1) = 2³²-1

=2³²-1

\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(=\dfrac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)

\(=\dfrac{5^{32}-1}{2}\)

\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{32}+1\right)\)

\(2P\)\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(2P=\)\(\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(2P=5^{32}-1\)

\(P=\dfrac{5^{32}-1}{2}\)