实数a,b,c,d满足：a＋b＋c＋d＝1001,ac＝bd＝4,则：求下列式子解根号下(a+b)(b+c)(c+d)(d+a)的值.

Question

实数a,b,c,d满足：a＋b＋c＋d＝1001,ac＝bd＝4,则：求下列式子解
根号下(a+b)(b+c)(c+d)(d+a) 的值.

Answer 1

=2002
因为 ac = bd = 4
则 abcd =4*4 =16
原式 =√(a+b)(b+c)(c+d)(d+a) =√[(a+b)(c+d)]*[(b+c)(d+a)]
=√(ac+ad+bc+bd)(ab+ac+bd+cd)
=√(8+16/bc+bc)(8+16/cd+cd)
=√[8bc+16+(bc)^2]/bc * [8cd+16+(cd)^2]/cd
=√(4+bc)^2/bc * (4+cd)^2/cd
=(4+bc)(4+cd)/2c
=(ac+bc)(4+4b/c)/2c
=2(a+b)(1+c/b)
=2(a+b+c+ac/b)
=2(a+b+c+d)
=2*1001
=2002