13.(16 分)计算:
(1)$ [\frac{7}{10} - (\frac{3}{5} - \frac{1}{3}) ÷ \frac{4}{5}] × \frac{2}{11} $
(2)$ (-2)^{3} × 5 - (-2.8) ÷ (-2)^{2} $
(1)$ [\frac{7}{10} - (\frac{3}{5} - \frac{1}{3}) ÷ \frac{4}{5}] × \frac{2}{11} $
(2)$ (-2)^{3} × 5 - (-2.8) ÷ (-2)^{2} $
答案
(1)$\frac{1}{15}$ (2)-39.3
解析
(1) $\left[\frac{7}{10} - \left(\frac{3}{5} - \frac{1}{3}\right) ÷ \frac{4}{5}\right] × \frac{2}{11}$
$=\left[\frac{7}{10} - \left(\frac{9}{15} - \frac{5}{15}\right) × \frac{5}{4}\right] × \frac{2}{11}$
$=\left[\frac{7}{10} - \frac{4}{15} × \frac{5}{4}\right] × \frac{2}{11}$
$=\left[\frac{7}{10} - \frac{1}{3}\right] × \frac{2}{11}$
$=\left[\frac{21}{30} - \frac{10}{30}\right] × \frac{2}{11}$
$=\frac{11}{30} × \frac{2}{11}$
$=\frac{1}{15}$
(2) $(-2)^{3} × 5 - (-2.8) ÷ (-2)^{2}$
$=-8 × 5 - (-2.8) ÷ 4$
$=-40 + 0.7$
$=-39.3$
14.(12 分)求代数式 $ 2(a^{2} - 2ab + 3b^{2}) - 3(-a^{2} + ab - 4b^{2}) $ 的值,其中 $ a = \frac{1}{3} $,$ b = -2 $.
答案
原式=$5a^{2}-7ab+18b^{2}$.当$a=\frac{1}{3},b=-2$时,原式=$77\frac{2}{9}$
解析
原式$=2a^{2}-4ab + 6b^{2}+3a^{2}-3ab + 12b^{2}$
$=5a^{2}-7ab + 18b^{2}$
当$a = \frac{1}{3}$,$b=-2$时,
原式$=5×(\frac{1}{3})^{2}-7×\frac{1}{3}×(-2)+18×(-2)^{2}$
$=5×\frac{1}{9}+\frac{14}{3}+18×4$
$=\frac{5}{9}+\frac{42}{9}+72$
$=\frac{47}{9}+72$
$=77\frac{2}{9}$
$=5a^{2}-7ab + 18b^{2}$
当$a = \frac{1}{3}$,$b=-2$时,
原式$=5×(\frac{1}{3})^{2}-7×\frac{1}{3}×(-2)+18×(-2)^{2}$
$=5×\frac{1}{9}+\frac{14}{3}+18×4$
$=\frac{5}{9}+\frac{42}{9}+72$
$=\frac{47}{9}+72$
$=77\frac{2}{9}$
15.(24 分)观察下列等式:$ \frac{1}{1 × 2} = 1 - \frac{1}{2} $,$ \frac{1}{2 × 3} = \frac{1}{2} - \frac{1}{3} $,$ \frac{1}{3 × 4} = \frac{1}{3} - \frac{1}{4} $.将以上三个等式两边分别相加,得 $ \frac{1}{1 × 2} + \frac{1}{2 × 3} + \frac{1}{3 × 4} = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} = 1 - \frac{1}{4} = \frac{3}{4} $.
(1)猜想并写出:$ \frac{1}{n(n + 1)} = $______;
(2)直接写出算式的结果:$ \frac{1}{1 × 2} + \frac{1}{2 × 3} + \frac{1}{3 × 4} + … + \frac{1}{205 × 206} = $______;
(3)探究并计算:$ \frac{1}{2 × 4} + \frac{1}{4 × 6} + \frac{1}{6 × 8} + … + \frac{1}{214 × 216} $.
(1)猜想并写出:$ \frac{1}{n(n + 1)} = $______;
(2)直接写出算式的结果:$ \frac{1}{1 × 2} + \frac{1}{2 × 3} + \frac{1}{3 × 4} + … + \frac{1}{205 × 206} = $______;
(3)探究并计算:$ \frac{1}{2 × 4} + \frac{1}{4 × 6} + \frac{1}{6 × 8} + … + \frac{1}{214 × 216} $.
答案
(1)$\frac{1}{n}-\frac{1}{n+1}$ (2)$\frac{205}{206}$ (3)因为$\frac{1}{2×4}=\frac{1}{4}×\frac{1}{1×2},\frac{1}{4×6}=\frac{1}{4}×\frac{1}{2×3},...$,所以原式=$\frac{1}{4}×(\frac{1}{1×2}+\frac{1}{2×3}+...+\frac{1}{107×108})=\frac{1}{4}×(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{107}-\frac{1}{108})=\frac{1}{4}×(1-\frac{1}{108})=\frac{1}{4}×\frac{107}{108}=\frac{107}{432}$
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