第三题没有答案吗?
追答(1)设x/(y+z)=y/(z+x)=z/(x+y)=k
那么x=k•(y+z)① y=k•(z+x)② z=k•(x+y)③
①+②+③:x+y+z=2k•(x+y+z)
∴k=1/2,x/(y+z)=k=1/2
(2)由a=2(b-1)=>a-2b=(-2)
∴2(a-2b)²+a²-4b²+a+6b
=2(a-2b)²+(a-2b)(a+2b)+(a-2b)+8b
=8-2(a+2b)-2+8b
=6-2a+4b
=6-2(a-2b)=6+4=10
(3)2x-y+z=0① 3x+y-2z=0②
①+②:z=5x ③ ③代入①y=7x
∵x•y•z≠0
∴(x-y+2z)/(2x+y-3z)=(x-7x+10x)/(2x+7x-15x)=(-2/3)