·ÖÎö£ºZÔ×ÓÓÐ2¸öµç×Ӳ㣬×îÍâ²ãµç×ÓÊýÊǺËÍâµç×ÓÊýµÄ
±¶£¬Áî×îÍâ²ãµç×ÓÊýΪa£¬Ôò
£¨2+a£©=a£¬½âµÃa=4£¬ËùÒÔZΪCÔªËØ£»
ÉèXÔ×ÓµÄÖÊ×ÓÊýΪx£¬YÔ×ÓµÄÖÊ×ÓÊýΪy£®
ÈýÖÖÔªËØÔ×ÓµÄÖÊ×Ó×ÜÊýΪ25£¬ÔòÓÐ6+x+y=25£»
ZºÍYµÄÔ×ÓÐòÊýÖ®ºÍ±ÈXµÄÔ×ÓÐòÊý2±¶»¹¶à1£¬ÔòÓÐ6+y=2x+1
ÁªÁ¢·½³Ì½âµÃ£ºx=8£¬y=11£¬¼´XΪÑõÔªËØ£¬YΪÄÆÔªËØ£®
È»ºóÀûÓõ¥Öʼ°»¯ºÏÎïµÄÐÔÖÊÀ´½â´ð£®
½â´ð£º½â£ºZÔ×ÓÓÐ2¸öµç×Ӳ㣬×îÍâ²ãµç×ÓÊýÊǺËÍâµç×ÓÊýµÄ
±¶£¬Áî×îÍâ²ãµç×ÓÊýΪa£¬Ôò
£¨2+a£©=a£¬½âµÃa=4£¬ËùÒÔZΪCÔªËØ£»
ÉèXÔ×ÓµÄÖÊ×ÓÊýΪx£¬YÔ×ÓµÄÖÊ×ÓÊýΪy£®
ÈýÖÖÔªËØÔ×ÓµÄÖÊ×Ó×ÜÊýΪ25£¬ÔòÓÐ6+x+y=25£»
ZºÍYµÄÔ×ÓÐòÊýÖ®ºÍ±ÈXµÄÔ×ÓÐòÊý2±¶»¹¶à1£¬ÔòÓÐ6+y=2x+1
ÁªÁ¢·½³Ì½âµÃ£ºx=8£¬y=11£¬¼´XΪÑõÔªËØ£¬YΪÄÆÔªËØ£®
£¨1£©ÓÉÉÏÊö·ÖÎö¿ÉÖª£¬XΪÑõÔªËØ£¬YΪÄÆÔªËØ£¬ZΪ̼ԪËØ£®
¹Ê´ð°¸Îª£ºÑõ£»ÄÆ£»Ì¼£®
£¨2£©ZX
2ÊÇCO
2£¬CO
2Êǹ²¼Û»¯ºÏÎ̼Ô×ÓÓëÑõÔ×ÓÖ®¼äÐγÉ2¶Ô¹²Óõç×Ó¶Ô£¬CO
2Ðγɹý³ÌΪ
¹Ê´ð°¸Îª£º
£®
£¨3£©Y
2XΪNa
2O£¬¶ÔӦˮ»¯ÎïΪNaOH£¬NaOHÊÇÀë×Ó»¯ºÏÎÓÉÄÆÀë×ÓÓëÇâÑõ¸ùÀë×Ó¹¹³É£¬ÇâÑõ»¯ÄƵç×ÓʽΪ
£®ÄÆÀë×ÓÓëÇâÑõ¸ùÀë×ÓÖ®¼äΪÀë×Ó¼ü£¬ÇâÑõ¸ùÀë×ÓÖÐÇâÔ×ÓÓëÑõÔ×ÓÖ®¼äΪ¹²¼Û¼ü£®
¹Ê´ð°¸Îª£º
£»Àë×Ó¼ü¡¢¹²¼Û¼ü£®
£¨4£©Y
2XΪNa
2O£¬ÈÜÓÚË®µÃNaOHÈÜÒº£¬ZX
2ÊÇCO
2£¬CO
2ÓëNaOHÈÜÒº·´Ó¦Éú³É̼ËáÄÆÓëË®£¬
·´Ó¦·½³ÌʽΪ2NaOH+CO
2=Na
2CO
3+H
2O£®
¹Ê´ð°¸Îª£º2NaOH+CO
2=Na
2CO
3+H
2O£®