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電子課本網(wǎng) 第166頁

第166頁

信息發(fā)布者:
解:?$(2)\begin {cases}3x + 2y=k + 1&①\\4x + 3y=k - 1&②\end {cases},$?
?$②×3-①×4$?得:?$y=-k - 7;$?
?$ ①×3-②×2$?得:?$x=k + 5。$?
?$ $?要使?$x,y$?均為連動數(shù),則?$-3\leq x\leq - 1$?或
?$1\leq x\leq 3,$??$-3\leq y\leq - 1$?或?$1\leq y\leq 3。$?
解得:?$-8\leq k\leq - 6$?或?$-4\leq k\leq - 2;$??
$-6\leq k\leq - 4$?或?$-10\leq k\leq - 8。$?
?$ $?所以?$k=-8$?或?$-6$?或?$-4。$?
?$ (3)$?解不等式組?$\begin {cases}\frac {2x - 6}{3}>x - 3\\\frac {x + 3}{2}\leq x - a\end {cases},$?
?$ $?解?$\frac {2x - 6}{3}>x - 3,$?解得?$x<3;$?
?$ $?解?$\frac {x + 3}{2}\leq x - a,$?解得?$x\geq 2a + 3。$?
?$ $?因為解集中恰好有?$4$?個連動整數(shù)解,
所以四個連動整數(shù)解為?$-2,-1,1,2,$?
所以?$-3<2a + 3\leq - 2,$?
?$ $?解得?$-3<a\leq -\frac {5}{2}。$?
∴?$a$?的取值范圍為?$-3<a\leq -\frac {5}{2}$?
$6$
$10$

解:任務(wù)二:設(shè)?$8$?個大小、形狀完全相同的小長方
形長為?$m,$?寬為?$n,$?
由題意得?$\begin {cases}m + 3n=13\\m + 2n-3n=5\end {cases},$?解得:?$\begin {cases}{m = 7}\\{n = 2}\end {cases}$?
則題圖②中陰影部分的面積為
?$13×(5 + 3×2)-8×7×2 = 31$?
任務(wù)三:由題意得
?$\begin {cases}2×18a + 3×18b + 4×18c + 144 = 36^2, \\2a + 2b + 2c = 36,\end {cases}$?
解得?$\begin {cases}b = 8 - 2a, \\c = a + 10.\end {cases}$?
∵?$a < b < c $?且?$a,$??$b,$??$c $?均為正整數(shù),
∴?$\begin {cases}a < 8 - 2a, \\8 - 2a < a + 10,\end {cases}$?解得?$0 < a < \frac {8}{3},$?
∴?$a = 1$?或?$2,$?
當(dāng)?$a = 2$?時,?$b = 8 - 2a = 4,$??$c = a + 10 = 12,$?
分別取?$A,$??$B,$??$C$?卡片?$3,$??$2,$??$5$?張,
拼成的不重疊的圖形面積為
?$3×18×2 + 2×18×4 + 5×18×12 $?
?$= 1332 > 36^2,$?故此時不能放置;
當(dāng)?$a = 1$?時,?$b = 8 - 2a = 6,$??$c = a + 10 = 11,$?
分別取?$A,$??$B,$??$C$?卡片?$3,$??$2,$??$5$?張,
拼成的不重疊的圖形面積為
?$3×18×1 + 2×18×6 + 5×18×11 $?
?$= 1260 < 36^2,$?故此時能放置,放置方式如圖。