设(a+b)(b+c)(c+d)(d+a)=(a+b+c+d)(bcd+cda+dab+abc),求证:ac=bd

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设(a+b)(b+c)(c+d)(d+a)=(a+b+c+d)(bcd+cda+dab+abc),求证:ac=bd设(a+b)(b+c)(c+d)(d+a)=(a+b+c+d)(bcd+cda+dab

设(a+b)(b+c)(c+d)(d+a)=(a+b+c+d)(bcd+cda+dab+abc),求证:ac=bd
设(a+b)(b+c)(c+d)(d+a)=(a+b+c+d)(bcd+cda+dab+abc),求证:ac=bd

设(a+b)(b+c)(c+d)(d+a)=(a+b+c+d)(bcd+cda+dab+abc),求证:ac=bd
(a + b) (b + c) (c + d) (d + a) - (a + b + c + d) (b c d + c d a + d a b + a b c)
=a^2c^2 - 2abcd + b^2d^2
=(ac-bd)^2
所以ac = bd

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