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$ \begin{aligned}解：原式&=x2-2×70x+702-702+4875 \\ &=(x-70)2-25 \\ &=(x-70)2-52 \\ &=(x-70+5)(x-70-5) \\ &=(x-65)(x-75) \\ \end{aligned}$

$ \begin{aligned}解：原式&=(2x)2-2×2x×1+12-12-575 \\ &=(2x-1)2-576 \\ &=(2x-1)2-242 \\ &=(2x-1+24)(2x-1-24) \\ &=(2x+23)(2x-25) \\ \end{aligned}$

$ \begin{aligned}解：原式&=x?+4x2y2+4y?-4x2y2 \\ &=(x2+2y2)2-(2xy)2 \\ &=(x2+2y2+2xy)(x2+2y2-2xy) \\ \end{aligned}$

$ \begin{aligned}解：原式&=x2-2ax+a2-a2-b2-2ab \\ &=(x-a)2-(a+b)2 \\ &=(x-a+a+b)(x-a-a-b) \\ &=(x+b)(x-2a-b) \\ \end{aligned}$

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$解：解法一(拆二次項(xiàng)):$

$ \begin{aligned}原式&=(x3+x2)+(x2-5x-6) \\ &=x2(x+1)+(x+ 1)(x-6) \\ &=(x+1)(x2+x-6) \\ &=(x+1)(x-2)(x+3) \\ \end{aligned}$

$解法二(拆一次項(xiàng)):$

$ \begin{aligned}原式&=(x3+2x2-8x)+(3x-6) \\ &=x(x2+2x-8)+3(x-2) \\ &=x(x-2)(x+4)+3(x-2) \\ &=(x-2)(x2+4x+3) \\ &=(x-2)(x+1)(x+3) \\ \end{aligned}$

$解法三(拆常數(shù)項(xiàng)):$

$ \begin{aligned}原式&=(x3+1)+(2x2-5x-7) \\ &=(x+1)(x2-x+1)+(x+1)(2x-7) \\ &=(x+1)(x2-x+1+2x-7) \\ &=(x+1)(x2+x-6) \\ &=(x+1)(x-2)(x+3) \\ \end{aligned}$

$解法四(拆二次項(xiàng)與一次項(xiàng)):$

$ \begin{aligned}原式&=(x3+x2)+(x2+x)-(6x+6) \\ &=x2(x+1)+x(x+1)-6(x+1) \\ &=(x+1)(x2+x-6) \\ &=(x+1)(x-2)(x+3) \\ \end{aligned}$

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