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2. (★★)计算：
(1) $\frac{1}{4} + (-\frac{2}{3}) + $
$ + (-\frac{1}{4}) + (-\frac{1}{3})$；
(2) $3.587 - (-5) + (-5\frac{1}{2}) + (+7) - (+3\frac{1}{4}) - (+1.587)$；
(3) $-0.125×7×(-5)×8$；
(4) $(-60)×(\frac{3}{4} + \frac{5}{6} - \frac{11}{15})$；
(5) $49\frac{24}{25}×(-5)$；
(6) $(-\frac{1}{12} + \frac{1}{3} - \frac{1}{2})÷(-\frac{1}{18})$；
(7) $25×\frac{3}{4} - (-25)×\frac{1}{2} + 25×(-\frac{1}{4})$；
(1) $\frac{1}{4} + (-\frac{2}{3}) + $
$ + (-\frac{1}{4}) + (-\frac{1}{3})$；(2) $3.587 - (-5) + (-5\frac{1}{2}) + (+7) - (+3\frac{1}{4}) - (+1.587)$；
(3) $-0.125×7×(-5)×8$；
(4) $(-60)×(\frac{3}{4} + \frac{5}{6} - \frac{11}{15})$；
(5) $49\frac{24}{25}×(-5)$；
(6) $(-\frac{1}{12} + \frac{1}{3} - \frac{1}{2})÷(-\frac{1}{18})$；
(7) $25×\frac{3}{4} - (-25)×\frac{1}{2} + 25×(-\frac{1}{4})$；
答案
解：
(1)
$\begin{aligned}原式&=[\frac{1}{4}+(-\frac{1}{4})]+[(-\frac{2}{3})+(-\frac{1}{3})]+\frac{5}{6}\\&=0+(-1)+\frac{5}{6}\\&=-\frac{1}{6}\end{aligned}$
(2)
$\begin{aligned}原式&=3.587 +5 -5\frac{1}{2} +7 -3\frac{1}{4} -1.587\\&=(3.587-1.587)+(5+7)+(-5\frac{1}{2}-3\frac{1}{4})\\&=2+12-8\frac{3}{4}\\&=5\frac{1}{4}\end{aligned}$
(3)
$\begin{aligned}原式&=(-0.125×8)×[7×(-5)]\\&=(-1)×(-35)\\&=35\end{aligned}$
(4)
$\begin{aligned}原式&=(-60)×\frac{3}{4}+(-60)×\frac{5}{6}-(-60)×\frac{11}{15}\\&=-45-50+44\\&=-51\end{aligned}$
(5)
$\begin{aligned}原式&=(50-\frac{1}{25})×(-5)\\&=50×(-5)-\frac{1}{25}×(-5)\\&=-250+\frac{1}{5}\\&=-249\frac{4}{5}\end{aligned}$
(6)
$\begin{aligned}原式&=(-\frac{1}{12}+\frac{1}{3}-\frac{1}{2})×(-18)\\&=(-\frac{1}{12})×(-18)+\frac{1}{3}×(-18)-\frac{1}{2}×(-18)\\&=\frac{3}{2}-6+9\\&=\frac{9}{2}\end{aligned}$
(7)
$\begin{aligned}原式&=25×\frac{3}{4}+25×\frac{1}{2}+25×(-\frac{1}{4})\\&=25×(\frac{3}{4}+\frac{1}{2}-\frac{1}{4})\\&=25×1\\&=25\end{aligned}$
(1)
$\begin{aligned}原式&=[\frac{1}{4}+(-\frac{1}{4})]+[(-\frac{2}{3})+(-\frac{1}{3})]+\frac{5}{6}\\&=0+(-1)+\frac{5}{6}\\&=-\frac{1}{6}\end{aligned}$
(2)
$\begin{aligned}原式&=3.587 +5 -5\frac{1}{2} +7 -3\frac{1}{4} -1.587\\&=(3.587-1.587)+(5+7)+(-5\frac{1}{2}-3\frac{1}{4})\\&=2+12-8\frac{3}{4}\\&=5\frac{1}{4}\end{aligned}$
(3)
$\begin{aligned}原式&=(-0.125×8)×[7×(-5)]\\&=(-1)×(-35)\\&=35\end{aligned}$
(4)
$\begin{aligned}原式&=(-60)×\frac{3}{4}+(-60)×\frac{5}{6}-(-60)×\frac{11}{15}\\&=-45-50+44\\&=-51\end{aligned}$
(5)
$\begin{aligned}原式&=(50-\frac{1}{25})×(-5)\\&=50×(-5)-\frac{1}{25}×(-5)\\&=-250+\frac{1}{5}\\&=-249\frac{4}{5}\end{aligned}$
(6)
$\begin{aligned}原式&=(-\frac{1}{12}+\frac{1}{3}-\frac{1}{2})×(-18)\\&=(-\frac{1}{12})×(-18)+\frac{1}{3}×(-18)-\frac{1}{2}×(-18)\\&=\frac{3}{2}-6+9\\&=\frac{9}{2}\end{aligned}$
(7)
$\begin{aligned}原式&=25×\frac{3}{4}+25×\frac{1}{2}+25×(-\frac{1}{4})\\&=25×(\frac{3}{4}+\frac{1}{2}-\frac{1}{4})\\&=25×1\\&=25\end{aligned}$
解析
【分析】
这7道题均为有理数四则运算题，解题时可通过观察式子特征，巧用运算律简化计算：加减运算中，优先将互为相反数、同分母分数、可凑整的数分组结合计算；乘除运算中，优先将乘积为整数的数结合，遇到带括号的运算或除法时，可转化为乘法后利用乘法分配律计算，减少硬算带来的错误。
【解析】
(1)
$\begin{aligned}原式&=[\frac{1}{4}+(-\frac{1}{4})]+[(-\frac{2}{3})+(-\frac{1}{3})]+\frac{5}{6}\\&=0+(-1)+\frac{5}{6}\\&=-\frac{1}{6}\end{aligned}$
(2)
$\begin{aligned}原式&=3.587 +5 -5\frac{1}{2} +7 -3\frac{1}{4} -1.587\\&=(3.587-1.587)+(5+7)+(-5\frac{1}{2}-3\frac{1}{4})\\&=2+12-8\frac{3}{4}\\&=5\frac{1}{4}\end{aligned}$
(3)
$\begin{aligned}原式&=(-0.125×8)×[7×(-5)]\\&=(-1)×(-35)\\&=35\end{aligned}$
(4)
$\begin{aligned}原式&=(-60)×\frac{3}{4}+(-60)×\frac{5}{6}-(-60)×\frac{11}{15}\\&=-45-50+44\\&=-51\end{aligned}$
(5)
$\begin{aligned}原式&=(50-\frac{1}{25})×(-5)\\&=50×(-5)-\frac{1}{25}×(-5)\\&=-250+\frac{1}{5}\\&=-249\frac{4}{5}\end{aligned}$
(6)
$\begin{aligned}原式&=(-\frac{1}{12}+\frac{1}{3}-\frac{1}{2})×(-18)\\&=(-\frac{1}{12})×(-18)+\frac{1}{3}×(-18)-\frac{1}{2}×(-18)\\&=\frac{3}{2}-6+9\\&=\frac{9}{2}\end{aligned}$
(7)
$\begin{aligned}原式&=25×\frac{3}{4}+25×\frac{1}{2}+25×(-\frac{1}{4})\\&=25×(\frac{3}{4}+\frac{1}{2}-\frac{1}{4})\\&=25×1\\&=25\end{aligned}$
【答案】
(1) $\boldsymbol{-\frac{1}{6}}$；(2) $\boldsymbol{5\frac{1}{4}}$；(3) $\boldsymbol{35}$；(4) $\boldsymbol{-51}$；(5) $\boldsymbol{-249\frac{4}{5}}$；(6) $\boldsymbol{\frac{9}{2}}$；(7) $\boldsymbol{25}$
【知识点】
有理数加减运算；有理数乘除运算；运算律简便计算
【点评】
本组题目是有理数运算的常考基础题型，核心考查加法、乘法运算律的灵活运用，熟练掌握分组凑整、拆分凑整、除法化乘法等技巧，可有效提升运算效率，降低符号错误概率，是后续代数运算的重要基础。
【难度系数】
0.7
这7道题均为有理数四则运算题，解题时可通过观察式子特征，巧用运算律简化计算：加减运算中，优先将互为相反数、同分母分数、可凑整的数分组结合计算；乘除运算中，优先将乘积为整数的数结合，遇到带括号的运算或除法时，可转化为乘法后利用乘法分配律计算，减少硬算带来的错误。
【解析】
(1)
$\begin{aligned}原式&=[\frac{1}{4}+(-\frac{1}{4})]+[(-\frac{2}{3})+(-\frac{1}{3})]+\frac{5}{6}\\&=0+(-1)+\frac{5}{6}\\&=-\frac{1}{6}\end{aligned}$
(2)
$\begin{aligned}原式&=3.587 +5 -5\frac{1}{2} +7 -3\frac{1}{4} -1.587\\&=(3.587-1.587)+(5+7)+(-5\frac{1}{2}-3\frac{1}{4})\\&=2+12-8\frac{3}{4}\\&=5\frac{1}{4}\end{aligned}$
(3)
$\begin{aligned}原式&=(-0.125×8)×[7×(-5)]\\&=(-1)×(-35)\\&=35\end{aligned}$
(4)
$\begin{aligned}原式&=(-60)×\frac{3}{4}+(-60)×\frac{5}{6}-(-60)×\frac{11}{15}\\&=-45-50+44\\&=-51\end{aligned}$
(5)
$\begin{aligned}原式&=(50-\frac{1}{25})×(-5)\\&=50×(-5)-\frac{1}{25}×(-5)\\&=-250+\frac{1}{5}\\&=-249\frac{4}{5}\end{aligned}$
(6)
$\begin{aligned}原式&=(-\frac{1}{12}+\frac{1}{3}-\frac{1}{2})×(-18)\\&=(-\frac{1}{12})×(-18)+\frac{1}{3}×(-18)-\frac{1}{2}×(-18)\\&=\frac{3}{2}-6+9\\&=\frac{9}{2}\end{aligned}$
(7)
$\begin{aligned}原式&=25×\frac{3}{4}+25×\frac{1}{2}+25×(-\frac{1}{4})\\&=25×(\frac{3}{4}+\frac{1}{2}-\frac{1}{4})\\&=25×1\\&=25\end{aligned}$
【答案】
(1) $\boldsymbol{-\frac{1}{6}}$；(2) $\boldsymbol{5\frac{1}{4}}$；(3) $\boldsymbol{35}$；(4) $\boldsymbol{-51}$；(5) $\boldsymbol{-249\frac{4}{5}}$；(6) $\boldsymbol{\frac{9}{2}}$；(7) $\boldsymbol{25}$
【知识点】
有理数加减运算；有理数乘除运算；运算律简便计算
【点评】
本组题目是有理数运算的常考基础题型，核心考查加法、乘法运算律的灵活运用，熟练掌握分组凑整、拆分凑整、除法化乘法等技巧，可有效提升运算效率，降低符号错误概率，是后续代数运算的重要基础。
【难度系数】
0.7
