第8页 - 通城学典课时作业本七年级数学苏科版江苏专用

$(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc$

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$(2a+b)(a+b)$

$解：(3)示意图不唯一，如图所示$

$(4)(a_{1}+a_{2})^2=a_{1}^2+2a_{1}a_{2}+a_{2}^2，共有1+2=3(项)；$

$(a_{1}+a_{2}+a_{3})^2=a_{1}^2+a_{2}^2+a_{3}^2+2a_{1}a_{2}+2a_{2}a_{3}+2a_{1}a_{3}$

$共有1+2+3=6(项)$

$∴将代数式(a_{1}+a_{2}+a_{3}+···+a_{20})^2$

$展开、合并同类项后共有1+2+3+···+20=\frac {(1+20)×20}{2}=210(项)$

$解：(1)原式=(a+b)²-2ab=3²-2×\frac {5}{4}=9-\frac {5}{2}=\frac {13}{2}$

$(2)(a-b)²=a²+b²-2ab =\frac {13}{2}-2×\frac {5}{4}=4$

$所以原式=±2$

$(3)由a+b=3，得a=3-b，代入ab=\frac {5}{4}\ $

$得(3-b)b= \frac {5}{4}，即-b²+3b=\frac {5}{4}$

$∴原式=2+2(-b²+3b) =2+2×\frac {5}{4}= \frac {9}{2}$