A )甲、乙两支同样质地的蜡烛、粗细、长短不同,甲支能燃3.5小时,乙支能燃5小时,燃了2小时后.两支蜡烛剩下之长度恰好相同.那么甲支与乙支蜡烛的长度之比为( )B)(1/11+1/21+1/31+1/41)*(1/21+1/31+
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![A )甲、乙两支同样质地的蜡烛、粗细、长短不同,甲支能燃3.5小时,乙支能燃5小时,燃了2小时后.两支蜡烛剩下之长度恰好相同.那么甲支与乙支蜡烛的长度之比为( )B)(1/11+1/21+1/31+1/41)*(1/21+1/31+](/uploads/image/z/10362337-25-7.jpg?t=A+%EF%BC%89%E7%94%B2%E3%80%81%E4%B9%99%E4%B8%A4%E6%94%AF%E5%90%8C%E6%A0%B7%E8%B4%A8%E5%9C%B0%E7%9A%84%E8%9C%A1%E7%83%9B%E3%80%81%E7%B2%97%E7%BB%86%E3%80%81%E9%95%BF%E7%9F%AD%E4%B8%8D%E5%90%8C%2C%E7%94%B2%E6%94%AF%E8%83%BD%E7%87%833.5%E5%B0%8F%E6%97%B6%2C%E4%B9%99%E6%94%AF%E8%83%BD%E7%87%835%E5%B0%8F%E6%97%B6%2C%E7%87%83%E4%BA%862%E5%B0%8F%E6%97%B6%E5%90%8E.%E4%B8%A4%E6%94%AF%E8%9C%A1%E7%83%9B%E5%89%A9%E4%B8%8B%E4%B9%8B%E9%95%BF%E5%BA%A6%E6%81%B0%E5%A5%BD%E7%9B%B8%E5%90%8C.%E9%82%A3%E4%B9%88%E7%94%B2%E6%94%AF%E4%B8%8E%E4%B9%99%E6%94%AF%E8%9C%A1%E7%83%9B%E7%9A%84%E9%95%BF%E5%BA%A6%E4%B9%8B%E6%AF%94%E4%B8%BA%28+%29B%29%EF%BC%881%2F11%2B1%2F21%2B1%2F31%2B1%2F41%29%2A%281%2F21%2B1%2F31%2B)
A )甲、乙两支同样质地的蜡烛、粗细、长短不同,甲支能燃3.5小时,乙支能燃5小时,燃了2小时后.两支蜡烛剩下之长度恰好相同.那么甲支与乙支蜡烛的长度之比为( )B)(1/11+1/21+1/31+1/41)*(1/21+1/31+
A )甲、乙两支同样质地的蜡烛、粗细、长短不同,甲支能燃3.5小时,乙支能燃5小时,燃了2小时后.两支蜡烛剩下之长度恰好相同.那么甲支与乙支蜡烛的长度之比为( )
B)(1/11+1/21+1/31+1/41)*(1/21+1/31+1/41+1/51)-(1/11+1/21+1/31+1/41+1/51)*(1/21+1/31+1/41)
A )甲、乙两支同样质地的蜡烛、粗细、长短不同,甲支能燃3.5小时,乙支能燃5小时,燃了2小时后.两支蜡烛剩下之长度恰好相同.那么甲支与乙支蜡烛的长度之比为( )B)(1/11+1/21+1/31+1/41)*(1/21+1/31+
A) 2/3.5x=2/5y x/y=7/10
B) 设1/11+1/21+1/31=a 1/21+1/31+1/41=b
原式化为(a+1/41)*(b+1/51)-(a+1/41+1/51)*b
化简得1/51(a-b)+1/(41x51)
综上,得:1/561
(1)假设甲长为3。5,已为5,那么一小时燃烧1,那么两个小时后,甲长为1。5,已为3,1。5不等于3,所以原来长度应该是2:1才相等
(2) 设1/11+1/21+1/31=a 1/21+1/31+1/41=b
原式化为(a+1/41)*(b+1/51)-(a+1/41+1/51)*b
化简得1/51(a-b)+1/(41x51)
综上,得:1/561...
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(1)假设甲长为3。5,已为5,那么一小时燃烧1,那么两个小时后,甲长为1。5,已为3,1。5不等于3,所以原来长度应该是2:1才相等
(2) 设1/11+1/21+1/31=a 1/21+1/31+1/41=b
原式化为(a+1/41)*(b+1/51)-(a+1/41+1/51)*b
化简得1/51(a-b)+1/(41x51)
综上,得:1/561
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1)
(3.5-2)甲=(5-2)乙
甲:乙=3:1.5=2:1
2)
(1/11+1/21+1/31+1/41)*(1/21+1/31+1/41+1/51)-(1/11+1/21+1/31+1/41+1/51)*(1/21+1/31+1/41)
设1/21+1/31+1/41=a
原式=(1/11+a)*(a+1/51)-(1/11+a+1/5...
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1)
(3.5-2)甲=(5-2)乙
甲:乙=3:1.5=2:1
2)
(1/11+1/21+1/31+1/41)*(1/21+1/31+1/41+1/51)-(1/11+1/21+1/31+1/41+1/51)*(1/21+1/31+1/41)
设1/21+1/31+1/41=a
原式=(1/11+a)*(a+1/51)-(1/11+a+1/51)*a
=(1/11a+a^2+1/51a+1/11×1/51)-(1/11a+a^2+1/51a)
=1/11×1/51
=1/561
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(1)假设甲长为3。5,已为5,那么一小时燃烧1,那么两个小时后,甲长为1。5,已为3,1。5不等于3,所以原来长度应该是2:1才相等
(2)A) 2/3.5x=2/5y x/y=7/10
B) 设1/11+1/21+1/31=a 1/21+1/31+1/41=b
原式化为(a+1/41)*(b+1/51)-(a+1/41+1/51)*b
化简得1/51(a...
全部展开
(1)假设甲长为3。5,已为5,那么一小时燃烧1,那么两个小时后,甲长为1。5,已为3,1。5不等于3,所以原来长度应该是2:1才相等
(2)A) 2/3.5x=2/5y x/y=7/10
B) 设1/11+1/21+1/31=a 1/21+1/31+1/41=b
原式化为(a+1/41)*(b+1/51)-(a+1/41+1/51)*b
化简得1/51(a-b)+1/(41x51)
综上,得:1/561
收起