設(shè),x支球隊(duì)
分析:
第一隊(duì)與除自己以外的球隊(duì)賽,所以要打(x-1)場(chǎng)
第二隊(duì)與除自己和第一隊(duì)以外的隊(duì)打,所以要打(x-2)場(chǎng)〔因?yàn)榕c第一隊(duì)的已經(jīng)計(jì)算過了〕
……
第x-1隊(duì)與前面各隊(duì)的比賽都已經(jīng)計(jì)算過了,還要與第x隊(duì)打1場(chǎng)
所以場(chǎng)次總共有:(x-1)+(x-2)+……+2+1
等差數(shù)列,首項(xiàng)(x-1),末項(xiàng)1,項(xiàng)數(shù)(x-1)項(xiàng),
求和公式:〔(x-1)+1 〕(x-1) / 2 =x(x-1) / 2
列方程:x(x-1)/2=30
化簡(jiǎn)為:x²-x-60=0
x支球隊(duì)參加籃球賽,參賽的每?jī)申?duì)之間都要比賽一場(chǎng),一共進(jìn)行了30場(chǎng)比賽,求參賽的籃球隊(duì)支數(shù)x.
x支球隊(duì)參加籃球賽,參賽的每?jī)申?duì)之間都要比賽一場(chǎng),一共進(jìn)行了30場(chǎng)比賽,求參賽的籃球隊(duì)支數(shù)x.
一元二次方程,求思路解析,只用列出算式.
一元二次方程,求思路解析,只用列出算式.
數(shù)學(xué)人氣:135 ℃時(shí)間:2020-04-26 15:56:11
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