Question

10£®¡°Éñ»úÃîË㡱¶ÔÓֿ죺£¨ÔõÑùËã¼ò±ã¾ÍÔõÑùË㣩
£¨1£©4.5¡Á3.7+4$\frac{1}{2}$¡Á1.3
 £¨2£©3¡Á£¨$\frac{2}{15}$+$\frac{1}{12}$£©-0.25
 £¨3£©1$\frac{1}{3}$+3.5-£¨$\frac{1}{4}$+$\frac{1}{12}$£©
£¨4£©11.3-£¨1.2-$\frac{11}{15}$£©+2$\frac{1}{3}$¡Á$\frac{1}{5}$£®
£¨5£©½â·½³Ì£º$\frac{2}{3}$x¡Â$\frac{7}{9}$=$\frac{9}{14}$
£¨6£©½â·½³Ì3y¡Á$\frac{9}{14}$=$\frac{3}{11}$£®

Answer 1

·ÖÎö £¨1£©°Ñ·ÖÊý±äСÊý£¬ÔÙ¸ù¾Ý³Ë·¨·ÖÅäÂɽøÐмòË㣻
£¨2£©Ïȸù¾Ý³Ë·¨·ÖÅäÂÉÈ¥À¨ºÅ£¬°ÑСÊý±ä·ÖÊý£¬ÔÙÀûÓüӷ¨½»»»ÂɺͽáºÏÂɼòË㣻
£¨3£©ÏÈËãÀ¨ºÅÀïµÄ¼Ó·¨£¬ÔÙÀûÓüӷ¨½»»»ÂɺͽáºÏÂɼòË㣬×îºóËã¼Ó·¨£»
£¨4£©ÏÈËã³Ë·¨£¬ÔÙ¸ù¾Ý¼õ·¨ÐÔÖÊÈ¥À¨ºÅ£¬×îºó¸ù¾Ý¼Ó·¨½»»»ÂɺͽáºÏÂɽøÐмòË㣻
£¨5£©¸ù¾ÝµÈʽÐÔÖÊ£¬·½³ÌÁ½±ßÏȳýÒÔ$\frac{7}{9}$£¬ÔÙ³ýÒÔ$\frac{2}{3}$½â´ð¼´¿É£»
£¨6£©¸ù¾ÝµÈʽÐÔÖÊ£¬·½³ÌÁ½±ßÏȳËÒÔ$\frac{14}{9}$£¬ÔÙ³ýÒÔ3½â´ð¼´¿É£®

½â´ð ½â£º£¨1£©4.5¡Á3.7+4$\frac{1}{2}$¡Á1.3
=4.5¡Á3.7+4.5¡Á1.3
=4.5¡Á£¨3.7+1.3£©
=4.5¡Á5
=22.5

£¨2£©3¡Á£¨$\frac{2}{15}$+$\frac{1}{12}$£©-0.25
=3¡Á$\frac{2}{15}$+3¡Á$\frac{1}{12}$-$\frac{1}{4}$
=$\frac{2}{5}$+£¨$\frac{1}{4}$-$\frac{1}{4}$£©
=$\frac{2}{5}$

£¨3£©1$\frac{1}{3}$+3.5-£¨$\frac{1}{4}$+$\frac{1}{12}$£©
=1$\frac{1}{3}$+3.5-$\frac{1}{3}$
=£¨1$\frac{1}{3}$-$\frac{1}{3}$£©+3.5
=1+3.5
=4.5

£¨4£©11.3-£¨1.2-$\frac{11}{15}$£©+2$\frac{1}{3}$¡Á$\frac{1}{5}$
=11.3-£¨1.2-$\frac{11}{15}$£©+$\frac{7}{15}$
=11.3-1.2+£¨$\frac{11}{15}$+$\frac{7}{15}$£©
=11.3-1.2+1.2
=11.3+£¨1.2-1.2£©
=11.3

£¨5£©$\frac{2}{3}$x¡Â$\frac{7}{9}$=$\frac{9}{14}$
   $\frac{2}{3}$x¡Â$\frac{7}{9}$¡Â$\frac{7}{9}$=$\frac{9}{14}$¡Â$\frac{7}{9}$
            $\frac{2}{3}$x=$\frac{81}{98}$
       $\frac{2}{3}$x$¡Â\frac{2}{3}$=$\frac{81}{98}$$¡Â\frac{2}{3}$
            x=$\frac{243}{196}$

£¨6£©3y¡Á$\frac{9}{14}$=$\frac{3}{11}$
  3y¡Á$\frac{9}{14}$¡Á$\frac{14}{9}$=$\frac{3}{11}$¡Á$\frac{14}{9}$
               3y=$\frac{14}{33}$
           3y¡Â3=$\frac{14}{33}$¡Â3
               y=$\frac{14}{99}$

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