若三个正数a,b,c成等比数列,三数x,y ,.z 成等差数列,则(y-z)lga+(z-x)lgb+(x-y)lgc=
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若三个正数a,b,c成等比数列,三数x,y ,.z 成等差数列,则(y-z)lga+(z-x)lgb+(x-y)lgc=
由题意: b^2 =ac , 2y = x + z 即:y-z = x-y
原式=lga^(y-z) + lgb^(z-x) + lgc^(x-y)
=lga^(x-y) + lgb^(z-x) + lgc^(x-y)
=lg[a^(x-y) × c^(x-y)] + lgb^(z-x)
=lg[(ac)^(x-y)] + lgb^(z-x)
=lg[(b^2)^(x-y)] + lgb^(z-x)
=lgb^(2x-2y) + lgb^(z-x)
=lgb^(2x-2y+z-x)
=lgb^(x-2y+z)
=lgb^[x-(x+z)+z]
=lgb^0
=lg1
=0
原式=lga^(y-z) + lgb^(z-x) + lgc^(x-y)
=lga^(x-y) + lgb^(z-x) + lgc^(x-y)
=lg[a^(x-y) × c^(x-y)] + lgb^(z-x)
=lg[(ac)^(x-y)] + lgb^(z-x)
=lg[(b^2)^(x-y)] + lgb^(z-x)
=lgb^(2x-2y) + lgb^(z-x)
=lgb^(2x-2y+z-x)
=lgb^(x-2y+z)
=lgb^[x-(x+z)+z]
=lgb^0
=lg1
=0
一.已知正数a,b,c成等比数列,x,y,z成等差数列,求证:(y-z)lga+(z-x)lgb+(x-y)lgc=0
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