分析 (1)根據(jù)加法交換律和結(jié)合律進(jìn)行簡算;
(2)根據(jù)減法的性質(zhì)進(jìn)行簡算;
(3)把64看作2×4×8,再根據(jù)乘法交換律和結(jié)合律進(jìn)行簡算;
(4)根據(jù)乘法分配律進(jìn)行簡算;
(5)分子=$\frac{5}{3}+\frac{11}{4}+\frac{19}{5}+…+\frac{9701}{99}+\frac{9899}{100}$,分母=$\frac{10}{3}+\frac{22}{4}+\frac{38}{5}+…+\frac{19402}{99}+\frac{19798}{100}$;分母中的數(shù)正好是分子的2倍,那么分母=($\frac{5}{3}+\frac{11}{4}+\frac{19}{5}+…+\frac{9701}{99}+\frac{9899}{100}$)×2,然后再化簡即可.
解答 解:(1)16.4+3.5+83.6+166.5
=(16.4+83.6)+(3.5+166.5)
=100+170
=270;
(2)9.16-5.72-1.28
=9.16-(5.72+1.28)
=9.16-7
=2.16;
(3)64×12.5×0.25×0.05
=(2×4×8)×12.5×0.25×0.05
=(2×0.05)×(4×0.25)×(8×12.5)
=0.1×1×100
=10;
(4)312.5×12.3-312.5×6.9+312.5
=312.5×(12.3-6.9+1)
=312.5×6.4
=(300+12.5)×6.4
=300×6.4+12.5×6.4
=1920+80
=2000;
(5)$\frac{1\frac{2}{3}+2\frac{3}{4}+3\frac{4}{5}+…+97\frac{98}{99}+98\frac{99}{100}}{3\frac{1}{3}+5\frac{2}{4}+7\frac{3}{5}+…+195\frac{97}{99}+197\frac{98}{100}}$
=$\frac{\frac{5}{3}+\frac{11}{4}+\frac{19}{5}+…+\frac{9701}{99}+\frac{9899}{100}}{\frac{10}{3}+\frac{22}{4}+\frac{38}{5}+…+\frac{19402}{99}+\frac{19798}{100}}$
=$\frac{\frac{5}{3}+\frac{11}{4}+\frac{19}{5}+…+\frac{9701}{99}+\frac{9899}{100}}{\frac{5}{3}×2+\frac{11}{4}×2+\frac{19}{5}×2+…+\frac{9701}{99}×2+\frac{9899}{100}×2}$
=$\frac{\frac{5}{3}+\frac{11}{4}+\frac{19}{5}+…+\frac{9701}{99}+\frac{9899}{100}}{(\frac{5}{3}+\frac{11}{4}+\frac{19}{5}+…+\frac{9701}{99}+\frac{9899}{100})×2}$
=$\frac{1}{2}$.
點評 考查了運算定律與簡便運算,四則混合運算.注意運算順序和運算法則,靈活運用所學(xué)的運算定律簡便計算.
科目:小學(xué)數(shù)學(xué) 來源: 題型:計算題
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科目:小學(xué)數(shù)學(xué) 來源: 題型:計算題
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