已知1/a,1/b,1/c,成等差数列,求证(b+c)/a,(a+c)/b,(a+b)/c也是等差数列
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![已知1/a,1/b,1/c,成等差数列,求证(b+c)/a,(a+c)/b,(a+b)/c也是等差数列](/uploads/image/z/13256583-15-3.jpg?t=%E5%B7%B2%E7%9F%A51%2Fa%2C1%2Fb%2C1%2Fc%2C%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E6%B1%82%E8%AF%81%EF%BC%88b%2Bc%29%2Fa%2C%28a%2Bc%29%2Fb%2C%28a%2Bb%29%2Fc%E4%B9%9F%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97)
已知1/a,1/b,1/c,成等差数列,求证(b+c)/a,(a+c)/b,(a+b)/c也是等差数列
已知1/a,1/b,1/c,成等差数列,求证(b+c)/a,(a+c)/b,(a+b)/c也是等差数列
已知1/a,1/b,1/c,成等差数列,求证(b+c)/a,(a+c)/b,(a+b)/c也是等差数列
1/a,1/b,1/c成等差数列
则 2/b=1/a+1/c
同时要证明(b+c)/a,(a+c)/b,(a+b)/c也成等差数列,
即证明
2*(a+c)/b=(b+c)/a+(a+b)/c
左边=(2/b)*(a+c)=(1/a+1/c)*(a+c)=2+a/c+c/a
右边=a/c+c/a+b*(1/a+1/c)=2+a/c+c/a
所以左边=右边
所以(b+c)/a,(a+c)/b,(a+b)/c成等差数列
1/a,1/b,1/c,成等差数列
2/b=1/a+1/c=(a+c)/ac
(b+c)/a+(a+b)/c
=(bc+c^2+a^2+ab)/ac
=[(a^2+c^2)+(ab+bc)]/ac
=[(a+c)^2-2ac+b(a+c)]/ac
=(a+c)^2/ac-2+b(a+c)/ac
由(a+c)/ac=2/b
所以上式=(a+c)*2/b-2+b*2/b
=2(a+c)/b-2+2
=2(a+c)/b
所以(b+c)/a,(a+c)/b,(a+b)/c也是等差数列
因为1/a,1/b,1/c,成等差数列,所以,2/b=1/a+1/c=(a+c)/ac,
即2ac=ab+bc
因为2(a+c)/b=(a+c)/ac×(a+c)=【a∧2+2ac+c∧2】/ac= [a∧2+ ab+bc+c∧2]/ac=[a(a+b)+b(b+c)]/ac=(b+c)/a+(a+b)/c
即证,(b+c)/a,(a+c)/b,(a+b)/c也是等差数列