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a^2+b^2=2c^2
c^2=1/2(a^2+b^2)
cosC=(a^2+b^2-c^2)/2ab=(a^2+b^2)/4ab>=2ab/4ab=1/2
最小值=1/2
CCCCCCCC
3)a^2+b^2+c^2=ab+bc+ac
2a^2+2b^2+2c^2=2ab+2bc+2ac
(a-b)^2+(b-c)^2+(a-c)^2=0
(a-b)^2=0, (b-c)^2=0, (a-c)^2=0
a=b-c
三角形ABC是等边三角形
3)x^2-y^2=2xy
等式两边同除以xy得
x/y-y/x=2
令x/y=t
t-1/t=2
t^2-2t-1=0
t=1+√2或t=1-√2
即,x/y=1+√2或t=1-√2
当x/y=1+√2有:
(x-y)/(x+y)=[x/y-1]/[x/y+1]=(1+√2-1)/(1+√2+1)=√2-1
当x/y=1-√2
(x-y)/(x+y)=[x/y-1]/[x/y+1]=(1-√2-1)/(1-√2+1)=-1-√2-
c^2=1/2(a^2+b^2)
cosC=(a^2+b^2-c^2)/2ab=(a^2+b^2)/4ab>=2ab/4ab=1/2
最小值=1/2
CCCCCCCC
3)a^2+b^2+c^2=ab+bc+ac
2a^2+2b^2+2c^2=2ab+2bc+2ac
(a-b)^2+(b-c)^2+(a-c)^2=0
(a-b)^2=0, (b-c)^2=0, (a-c)^2=0
a=b-c
三角形ABC是等边三角形
3)x^2-y^2=2xy
等式两边同除以xy得
x/y-y/x=2
令x/y=t
t-1/t=2
t^2-2t-1=0
t=1+√2或t=1-√2
即,x/y=1+√2或t=1-√2
当x/y=1+√2有:
(x-y)/(x+y)=[x/y-1]/[x/y+1]=(1+√2-1)/(1+√2+1)=√2-1
当x/y=1-√2
(x-y)/(x+y)=[x/y-1]/[x/y+1]=(1-√2-1)/(1-√2+1)=-1-√2-
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请问最后一题(线性规划)怎么做?谢谢
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