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1. 计算:
(1)$(-2)^{3}+\dfrac{1}{4}×[1-(-3)^{2}]$;
(2)$-3^{2}+|5-8|+24÷(-3)×\dfrac{1}{3}$.
(1)$(-2)^{3}+\dfrac{1}{4}×[1-(-3)^{2}]$;
(2)$-3^{2}+|5-8|+24÷(-3)×\dfrac{1}{3}$.
答案
(1)-10 (2)$-\dfrac{26}{3}$
2. 计算:
(1) $[4\dfrac{2}{3} × (-\dfrac{5}{14}) + (-0.4) ÷ (-\dfrac{4}{25})] × 1\dfrac{1}{5}$；
(2) $-2^2 + (-2)^2 - (-1)^3 × (\dfrac{1}{3} - \dfrac{1}{2}) ÷ \dfrac{1}{6} - |-1|$；
(3) $(-3)^3 + (-3) × [(-4)^2 + 2] - (-4)^2 ÷ (-2)$；
(4) $\dfrac{1}{1 × 2} + \dfrac{1}{2 × 3} + \dots + \dfrac{1}{98 × 99}$。
(1) $[4\dfrac{2}{3} × (-\dfrac{5}{14}) + (-0.4) ÷ (-\dfrac{4}{25})] × 1\dfrac{1}{5}$；
(2) $-2^2 + (-2)^2 - (-1)^3 × (\dfrac{1}{3} - \dfrac{1}{2}) ÷ \dfrac{1}{6} - |-1|$；
(3) $(-3)^3 + (-3) × [(-4)^2 + 2] - (-4)^2 ÷ (-2)$；
(4) $\dfrac{1}{1 × 2} + \dfrac{1}{2 × 3} + \dots + \dfrac{1}{98 × 99}$。
答案
(1)1
(2)-2
(3)-73
(4)原式$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dots+\dfrac{1}{98}-\dfrac{1}{99}=1-\dfrac{1}{99}=\dfrac{98}{99}$.
【错因分析】很多同学对于这种题目不知如何下手，一般地,根据$\dfrac{1}{ab}=\dfrac{1}{(b-a)}×(\dfrac{1}{a}-\dfrac{1}{b})$对式子变形,注意$\dfrac{1}{1×2}=1-\dfrac{1}{2}$与$\dfrac{1}{3×5}=\dfrac{1}{2}×(\dfrac{1}{3}-\dfrac{1}{5})$这两类变形的区别.
(2)-2
(3)-73
(4)原式$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dots+\dfrac{1}{98}-\dfrac{1}{99}=1-\dfrac{1}{99}=\dfrac{98}{99}$.
【错因分析】很多同学对于这种题目不知如何下手，一般地,根据$\dfrac{1}{ab}=\dfrac{1}{(b-a)}×(\dfrac{1}{a}-\dfrac{1}{b})$对式子变形,注意$\dfrac{1}{1×2}=1-\dfrac{1}{2}$与$\dfrac{1}{3×5}=\dfrac{1}{2}×(\dfrac{1}{3}-\dfrac{1}{5})$这两类变形的区别.
