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標題:
Maths(Urgent !!)--20pt
發問:
--------------------------------------------------------------------------------1= 1 + bcb= b + bdc = c + dcd=cb + d^2Whts are all the possible values of b, c and d?Do the ans are:b= 0 , c = 0 , d=0 ; orb=0, d= 1, c can be any real no. ; orb can be any real no., c=0, d=1correct?any mistakes writing?THZ 顯示更多 -------------------------------------------------------------------------------- 1= 1 + bc b= b + bd c = c + dc d=cb + d^2 Whts are all the possible values of b, c and d? Do the ans are: b= 0 , c = 0 , d=0 ; or b=0, d= 1, c can be any real no. ; or b can be any real no., c=0, d=1 correct?any mistakes writing? THZ
最佳解答:
The answer should be: b = 0, c = 0, d = 1; OR b = 0, c can be any real number, d = 0; OR b can be any real number, c= 0, d = 0. 1 = 1 + bc............(1) b = b + bd...........(2) c = c + dc............(3) d = cb + d^2.......(4) From (1), (2), (3), we have bc = 0, bd = 0, cd = 0 Then as bc = 0, from (4), we have d = 0 + d^2 d^2 - d = 0 d(d - 1) = 0 d = 0 or d = 1 Case 1: d = 1. Then as cd = 0, we have c = 0. Also as bd = 0, we have b = 0. Case 2: d = 0. Then we only have bc = 0. If b = 0, then c can be any real number. If c = 0, then b can be any real number. CONCLUSION: b = 0, c = 0, d = 1; OR b = 0, c can be any real number, d = 0; OR b can be any real number, c= 0, d = 0.
其他解答:
d=cb + d^2 知道d=1 因為cb=0 (從第一式知道) d=d^2 d must be equal to 1 PUT d= 1 into b= b + bd so that we get b = b + b so b must be equal 0 then in 1= 1 + bc & c = c + dc PUT b = 0 and d = 1 1 = 1 + 0 correct but c = c + c so b must be equal 0 again so the mistake may be c = c + dc should be +bc
Maths(Urgent !!)--20pt
發問:
--------------------------------------------------------------------------------1= 1 + bcb= b + bdc = c + dcd=cb + d^2Whts are all the possible values of b, c and d?Do the ans are:b= 0 , c = 0 , d=0 ; orb=0, d= 1, c can be any real no. ; orb can be any real no., c=0, d=1correct?any mistakes writing?THZ 顯示更多 -------------------------------------------------------------------------------- 1= 1 + bc b= b + bd c = c + dc d=cb + d^2 Whts are all the possible values of b, c and d? Do the ans are: b= 0 , c = 0 , d=0 ; or b=0, d= 1, c can be any real no. ; or b can be any real no., c=0, d=1 correct?any mistakes writing? THZ
最佳解答:
The answer should be: b = 0, c = 0, d = 1; OR b = 0, c can be any real number, d = 0; OR b can be any real number, c= 0, d = 0. 1 = 1 + bc............(1) b = b + bd...........(2) c = c + dc............(3) d = cb + d^2.......(4) From (1), (2), (3), we have bc = 0, bd = 0, cd = 0 Then as bc = 0, from (4), we have d = 0 + d^2 d^2 - d = 0 d(d - 1) = 0 d = 0 or d = 1 Case 1: d = 1. Then as cd = 0, we have c = 0. Also as bd = 0, we have b = 0. Case 2: d = 0. Then we only have bc = 0. If b = 0, then c can be any real number. If c = 0, then b can be any real number. CONCLUSION: b = 0, c = 0, d = 1; OR b = 0, c can be any real number, d = 0; OR b can be any real number, c= 0, d = 0.
其他解答:
d=cb + d^2 知道d=1 因為cb=0 (從第一式知道) d=d^2 d must be equal to 1 PUT d= 1 into b= b + bd so that we get b = b + b so b must be equal 0 then in 1= 1 + bc & c = c + dc PUT b = 0 and d = 1 1 = 1 + 0 correct but c = c + c so b must be equal 0 again so the mistake may be c = c + dc should be +bc
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