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第37頁(yè)

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$解：原式?=\frac 25÷(-\frac {12}{5})-\frac {8}{21}×(-\frac {7}{4})-\frac 14?$

$?=\frac 25×(-\frac {5}{12})+\frac 8{21}×\frac 74-\frac 14?$

$?=-\frac 16+\frac 23-\frac 14?$

$?=\frac 12-\frac 14?$

$?=\frac {1}{4}?$

$解：原式?=-4×(-\frac {3}{2})-32÷4×(-\frac {5}{4})?$

$?=4×\frac 32+32×\frac 14×\frac 54?$

$?=6+10?$

$?=16?$

$解：?(\frac 16-\frac 27+\frac 23-\frac {3}{14})÷(-\frac {1}{42})?$

$?=(\frac 16-\frac 27+\frac 23-\frac {3}{14})×(-42)?$

$?=-7+12-28+9?$

$?=-14?$

$所以?(-\frac {1}{42})÷(\frac 16-\frac 27+\frac 23-\frac {3}{14})=-\frac {1}{14} .?$

$解：因?yàn)?|a|=4?，?|b|=9?，?|c|=6?，所以?a=±4?，?b=±9?，?c=±6.?$

$因?yàn)?ab＜0?，?bc＞0?，$

$所以?a=4?時(shí)，?b=-9?，?c=-6?或?a=-4?時(shí)，?b=9?，?c=6.?$

$所以?ab-bc+abc=4×(-9)-(-9)×(-6)+4×(-9)×(-6)=126?$

$或?ab-bc+abc=-4×9-9×6+(-4)×9×6=-306?$

$故?ab-bc+abc ?的值為?126?或?-306.?$

$?1-\frac {1}{2^n}?$

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