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第56页

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$ \begin{aligned} 解：原式&=(m^3)^2-1^2 \\ &=m^6-1. \\ \end{aligned}$

$ \begin{aligned}解：原式&=4x²+12xy+9y²-(4x²-12xy+9y²) \\ &=24xy. \\ \end{aligned}$

$ \begin{aligned}解：原式&=a²+2ab+b²-(a²-b²) \\ &=2ab+2b². \\ \end{aligned}$

$ \begin{aligned}解：原式&=4x²+12x+9-(4x²-9) \\ &=4x²+12x+9-4x²+9 \\ &=12x+18. \\ \end{aligned}$

$解：原式=a²+4a+4+a²-9-2a²-ab=4a-ab-5，$

$因为\frac{1}{2}ab=2a+1，$

$所以ab=4a+2. $

$所以原式=4a-(4a+2)-5=-7.$

$ \begin{aligned} 解：原式&=2(x+6)(x-6) \\ &=2(x²-36) \\ &=2x²-72. \\ \end{aligned}$

$ $

$ \begin{aligned} 解：原式&=[(3y-2z)+(4x+1)][(3y-2z)-(4x+1)] \\ &=(3y-2z)²-(4x+1)² \\ &=(9y²-12yz+4z²)-(16x²+8x+1) \\ &=9y²-12yz+4z²-16x²-8x-1. \\ \end{aligned}$

$ $

$ \begin{aligned} 解：原式&=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)·(2^{32}+1)(2^{64}+1)(2^{128}+1)+1 \\ &=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)(2^{64}+1)·(2^{128}+1)+1 \\ \end{aligned}$

$\quad \quad \quad \quad ···$

$ \begin{aligned} \quad \quad \quad \quad &=2^{256}-1+1 \\ \quad \quad \quad \quad &=2^{256}. \\ \end{aligned}$