第23页 - 通城学典课时作业本七年级数学苏科版江苏专用

A

1

−5

−9

8

$ \begin{aligned}解：原式&=\frac {1}{4}a²-2×\frac {1}{2}a×2b+(2b)² \\ &=\frac {1}{4}a²-2ab+4b² \\ \end{aligned}$

$ \begin{aligned}解：原式&=x³+x²-x-1-(x³-4x²-2x+8) \\ &=x³+x²-x-1-x³+4x²+2x-8 \\ &=5x²+x-9 \\ \end{aligned}$

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

$ \begin{aligned}解：原式&=6x²+17x+12-2(x²-3x+2) \\ &=6x²+17x+12-2x²+6x-4 \\ &=4x²+23x+8 \\ \end{aligned}$

$解：S=\frac {1}{2}×(5a+2b+4a+3b)×(2a+b)$

$=\frac {1}{2}×(9a+5b)×(2a+b)$

$=\frac {1}{2}×(18a²+9ab+10ab+5b²)$

$=(9a²+\frac {19}{2}ab+\frac {5}{2}b²)\mathrm {cm}²$