分析:先根據(jù)分?jǐn)?shù)除法的計(jì)算方法,變成乘法后約分,再根據(jù)比例的性質(zhì)進(jìn)得出兩個(gè)未知數(shù)的積,然后討論求解即可.
解答:解:(1)設(shè)兩個(gè)未知數(shù)是a和b,那么:
÷12=
×
=
=
;
2ab=14,
ab=7;
只要a與b的積是7即可;
①當(dāng)a=7,b=1時(shí);
即:
÷12=
;
②當(dāng)a=1,b=7時(shí):
÷12=
;
任取一組即可.
(2)設(shè)兩個(gè)未知數(shù)分別是a和b,那么:
÷3=
×
=
=
;
因?yàn)?span id="11cy6jg" class="MathJye" mathtag="math" style="whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal">
=
;
所以:ab=33×7=11×21=77×3=231×1;
①當(dāng)a=33,b=7時(shí):
÷3=
;
②當(dāng)a=7,b=33時(shí):
÷3=
;
③當(dāng)a=11,b=21時(shí);
÷3=
;
④當(dāng)a=21,b=11時(shí):
÷3=
;
⑤當(dāng)a=77,b=3時(shí):
÷3=
;
⑥當(dāng)a=3,b=77時(shí):
÷3=
;
⑦當(dāng)a=231,b=1時(shí):
=
;
⑧當(dāng)a=1,b=231時(shí):
÷3=
.
只要任取其中一組即可.
故答案為:7,1;33,7.(答案不唯一).
點(diǎn)評(píng):本題先根據(jù)分?jǐn)?shù)除法的計(jì)算方法,進(jìn)行化簡(jiǎn),根據(jù)比例的性質(zhì)找出兩個(gè)未知的積,再討論求解即可.