C = (2x-3)²-(x+4)(2x-1) -(x+3)²

(Chuyển đổi các hằng đẳng thức)

    = (4x²-12x+9)-(2x²-x+8x-4)-(x²+6x+9)

    =  4x²-12x+9-2x²+x-8x+4-x²-6x-9

(Ta thu gọn các hạng tử đồng dạng với nhau)

    = x²-25x-14 

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

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

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

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

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

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

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

a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

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

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

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

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

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

b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)

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

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

\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)

bạn ơi!! 2⁸ +1 chứ, bn xem lại đề cấy

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

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

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

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

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

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

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

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

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

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

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

=>  \(A=\frac{2^{32}-1}{3}\)

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

Đặt A=3(22 +1)(24+1)(28+1)(216+1)

=(4-1)(2^2+1)(2^4+1)(28+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)

Theo mình ý a bn làm đc