△ABC三边abc和面积满足S=c2-(a-b)2,且a+b=2△ABC的三边a,b,c和面积S满足S=c2-(a-b)2,且a+b=2,求面积S的最大值面积公式:S=1/2ab*sinC和余弦定理如下:S=(absinC)/2c^2-(a-b)^2=c^2-a^2-b^2+2ab=2ab(1-cosC)得sinC=4(1-cosC)
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![△ABC三边abc和面积满足S=c2-(a-b)2,且a+b=2△ABC的三边a,b,c和面积S满足S=c2-(a-b)2,且a+b=2,求面积S的最大值面积公式:S=1/2ab*sinC和余弦定理如下:S=(absinC)/2c^2-(a-b)^2=c^2-a^2-b^2+2ab=2ab(1-cosC)得sinC=4(1-cosC)](/uploads/image/z/14148768-48-8.jpg?t=%E2%96%B3ABC%E4%B8%89%E8%BE%B9abc%E5%92%8C%E9%9D%A2%E7%A7%AF%E6%BB%A1%E8%B6%B3S%3Dc2-%28a-b%292%2C%E4%B8%94a%2Bb%3D2%E2%96%B3ABC%E7%9A%84%E4%B8%89%E8%BE%B9a%2Cb%2Cc%E5%92%8C%E9%9D%A2%E7%A7%AFS%E6%BB%A1%E8%B6%B3S%3Dc2-%28a-b%EF%BC%892%2C%E4%B8%94a%2Bb%3D2%2C%E6%B1%82%E9%9D%A2%E7%A7%AFS%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E9%9D%A2%E7%A7%AF%E5%85%AC%E5%BC%8F%EF%BC%9AS%3D1%2F2ab%2AsinC%E5%92%8C%E4%BD%99%E5%BC%A6%E5%AE%9A%E7%90%86%E5%A6%82%E4%B8%8B%EF%BC%9AS%3D%28absinC%29%2F2c%5E2-%28a-b%29%5E2%3Dc%5E2-a%5E2-b%5E2%2B2ab%3D2ab%281-cosC%29%E5%BE%97sinC%3D4%281-cosC%29)
△ABC三边abc和面积满足S=c2-(a-b)2,且a+b=2△ABC的三边a,b,c和面积S满足S=c2-(a-b)2,且a+b=2,求面积S的最大值面积公式:S=1/2ab*sinC和余弦定理如下:S=(absinC)/2c^2-(a-b)^2=c^2-a^2-b^2+2ab=2ab(1-cosC)得sinC=4(1-cosC)
△ABC三边abc和面积满足S=c2-(a-b)2,且a+b=2
△ABC的三边a,b,c和面积S满足S=c2-(a-b)2,且a+b=2,求面积S的最大值
面积公式:S=1/2ab*sinC
和余弦定理
如下:
S=(absinC)/2
c^2-(a-b)^2=c^2-a^2-b^2+2ab=2ab(1-cosC)
得sinC=4(1-cosC),两边平方后
1-(cosC)^2=16(1-cosC)^2
(1-cosC)(15+17cosC)=0
cosC=-15/17 (cosC=1时C=0,舍去)
sinC=8/17
S最大值为S=(absinC)/2≤4/17
这道题目网上有解答,但是“由a+b≥2根号(ab)得ab≤1”倒数第二步,是怎么得到这个式子的?
△ABC三边abc和面积满足S=c2-(a-b)2,且a+b=2△ABC的三边a,b,c和面积S满足S=c2-(a-b)2,且a+b=2,求面积S的最大值面积公式:S=1/2ab*sinC和余弦定理如下:S=(absinC)/2c^2-(a-b)^2=c^2-a^2-b^2+2ab=2ab(1-cosC)得sinC=4(1-cosC)
a+b=2
由a+b≥2√ab
2≥2√ab
√ab≤1
ab≤1
所以
sinC=8/17
S最大值为S=(absinC)/2≤(1x8/17)/2=4/17