已知a+2b-3c/2=b-2c+3a/3=c+3a+2b/4,求a-2b+3c/a+3b-2c的值
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已知a+2b-3c/2=b-2c+3a/3=c+3a+2b/4,求a-2b+3c/a+3b-2c的值
已知a+2b-3c/2=b-2c+3a/3=c+3a+2b/4,求a-2b+3c/a+3b-2c的值
已知a+2b-3c/2=b-2c+3a/3=c+3a+2b/4,求a-2b+3c/a+3b-2c的值
楼上的做法是在k不等于零的情况下才成立.个人认为正确的做法是:先通分,然后三个式子两两相减,得3a-4b+5c=0 a-2b-7c=0 3a-2b-11c=0 然后把a和b都化成关于c的等式,带入要求的式子即可.结果是2/9.手机打的,所以不太详细.
令(a+2b-3c)/2=(b-2c+3a)/3=(c+3a+2b)/4=k,得
a+2b-3c=2k...①
3a+b-2c=3k...②
3a+2b+c=4k...③
③-②得b+3c=k④
④+①得a+3b=3k⑤
③-①得a+2c=k⑥
解得a=9k/11,b=8k/11,c=k/11,
带入(a-2b+3c)/(a+3b-2c)
=(9k/11-16k/11+3k/11)/(9k/11+24k/11-2k/11)
=-4/31