第80页 - 经纶学典学霸七年级数学上下册【江苏国标】

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6

b＜c＜a

$ \begin{aligned}解：原式&=4x²-4xy+y²-x²+y² \\ &=3x²-4xy+2y² \\ \end{aligned}$

$ \begin{aligned}解：原式&=[(3x+y)(3x-y)]² \\ &=(9x²-y²)² \\ &=81x⁴-18x²y²+y⁴ \\ \end{aligned}$

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

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

$ \begin{aligned}解：原式&=(x-y)(9-m²) \\ &=(x-y)(3+m)(3-m) \\ \end{aligned}$

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

$ \begin{aligned}解：原式&=(y²-1)²-6(y²-1)+9 \\ &=[(y²-1)-3]² \\ &=(y²-4)² \\ &=[(y+2)(y-2)]² \\ &=(y+2)²(y-2)² \\ \end{aligned}$

$ \begin{aligned}解：原式&=9x²-4y²-4xy+4y² \\ &=9x²-4xy \\ \end{aligned}$

$当x=\frac{1}{3}，y=-2时$

$原式= 9×(\frac{1}{3})²-4×\frac{1}{3}×(-2)=\frac{11}{3}$

$ \begin{aligned}解：原式&=x²-6x+9+2x²+14x-4x-28-x²+4 \\ &=2x²+4x-15 \\ \end{aligned}$

$因为x²+2x-4=0，所以x²+2x=4$

$当x²+2x=4时$

$原式=2(x²+2x)-15=2×4-15=-7.$