YLL討論網

san 寫到:#ed_op#DIV#ed_cl##ed_op#P#ed_cl#1.(x^3+y^3)/(x^3-y^3)=-234/109#ed_op#/P#ed_cl##ed_op#P#ed_cl#求(x^2+y^2)/(x^2-y^2)=?#ed_op#/P#ed_cl##ed_op#/DIV#ed_cl#

#ed_op#DIV#ed_cl#另一種解法#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#x#ed_op#SUP#ed_cl#3#ed_op#/SUP#ed_cl#+y#ed_op#SUP#ed_cl#3#ed_op#/SUP#ed_cl#=234k(k不等於0)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#y#ed_op#SUP#ed_cl#3#ed_op#/SUP#ed_cl#-x#ed_op#SUP#ed_cl#3#ed_op#/SUP#ed_cl#=109k#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#2x#ed_op#SUP#ed_cl#3#ed_op#/SUP#ed_cl#=125k → x=5*(k/2)#ed_op#SUP#ed_cl#1/3#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#2y#ed_op#SUP#ed_cl#3#ed_op#/SUP#ed_cl#=343k → y=7*(k/2)#ed_op#SUP#ed_cl#1/3#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+y#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#)/(x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#-y#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#)=[25*(k/2)#ed_op#SUP#ed_cl#2/3#ed_op#/SUP#ed_cl#]+[49*(k/2)#ed_op#SUP#ed_cl#2/3#ed_op#/SUP#ed_cl#]/[25*(k/2)#ed_op#SUP#ed_cl#2/3#ed_op#/SUP#ed_cl#]-[49*(k/2)#ed_op#SUP#ed_cl#2/3#ed_op#/SUP#ed_cl#]#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=74/(-24)=-37/12#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl#

san 寫到:#ed_op#div#ed_cl##ed_op#p#ed_cl#1.(x^3+y^3)/(x^3-y^3)=-234/109#ed_op#/p#ed_cl##ed_op#p#ed_cl#求(x^2+y^2)/(x^2-y^2)=?#ed_op#/p#ed_cl##ed_op#p#ed_cl# #ed_op#/p#ed_cl##ed_op#p#ed_cl#2.2x^4-3x^3+5x^2-3x+2=0#ed_op#/p#ed_cl##ed_op#p#ed_cl#求x+  1/x=?#ed_op#/p#ed_cl##ed_op#p#ed_cl# #ed_op#/p#ed_cl##ed_op#/div#ed_cl#

#ed_op#br#ed_cl#1. (x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#+y#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#)/(x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#-y#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#) = -(y#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#+x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#)/(y#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#-x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#) = -(y#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#-x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#+2x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#)/(y#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#-x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#) = -[1+ 2x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#/(y#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#-x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#)] = -234/109 = - (1+ 125/109)#ed_op#br#ed_cl#=> 2x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#/(y#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#-x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#) = 125/109 => 218x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl# = 125y#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#-125x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl# => 343x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#=125y#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl# => 7x=5y => y = (7/5)x#ed_op#br#ed_cl#(x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+y#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#)/(x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-y#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#) = (x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+(7/5)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#)/(x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-(7/5)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#) = (1+49/25)x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#/(1-49/25)x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl# = (74/25)/(-24/5) = -37/12#ed_op#br#ed_cl##ed_op#br#ed_cl#2. 設 y= x+1/x，y#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl# = (x+1/x)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl# = x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+2+1/x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#;#ed_op#br#ed_cl#0 = 2x#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#-3x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#+5x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-3x+2，若x=0，則0=2 =>故x=/=0，將兩邊除以x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#得：#ed_op#br#ed_cl#=> 0 = 2x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-3x+5-3/x+2/x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl# = 2x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+4+2/x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl# - 3x-3/x + 1 = 2(x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+2+1/x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#) - 3(x+1/x) +1 = 2y#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-3y+1 = (2y-1)(y-1) => y = 1/2 or 1#ed_op#br#ed_cl#驗證： 1/2=x+1/x or 1=x+1/x => x/2=x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+1 or x=x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+1 => 2x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-x+2=0 or x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-x+1=0#ed_op#br#ed_cl#b#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-4ac = (-1)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-4(2)(2) or (-1)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-4(1)(1)=-15or-3<0，故若x為實數，則此題無解；若x能為虛數，則 x+1/x = 1/2 或 1。#ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#sup#ed_cl##ed_op#/sup#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl#

#ed_op#DIV#ed_cl##ed_op#P#ed_cl#1.(x^3+y^3)/(x^3-y^3)=-234/109#ed_op#/P#ed_cl##ed_op#P#ed_cl#求(x^2+y^2)/(x^2-y^2)=?#ed_op#/P#ed_cl##ed_op#P#ed_cl# #ed_op#/P#ed_cl##ed_op#P#ed_cl#2.2x^4-3x^3+5x^2-3x+2=0#ed_op#/P#ed_cl##ed_op#P#ed_cl#求x+  1/x=?#ed_op#/P#ed_cl##ed_op#P#ed_cl# #ed_op#/P#ed_cl##ed_op#/DIV#ed_cl#