①若a\b=b\c=c\d=d\a,求a-b+c-d\a+b-c+d.
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①若a\b=b\c=c\d=d\a,求a-b+c-d\a+b-c+d.
②已知x,y,z互不相等,且x+1\y=y+1\z=z+1\x,求证x²y²z²=1.
③已知x,y,z互不相等,且x+2\y=y+2\z=z+2\x=K,求K的值.
④已知x.y.z满足x+1\y=4,y+1\z=1,z+1\x=7\3,求xyz的值.
⑤已知a+b+c=0,且1\(a+1)+1\(b+2)+1\(c+3)=0,求(a+1)²+(b+2)²+(c+3)²的值.
sorry 再加两题①已知x²-4x+1=0,求x²\x四次方+x²+1.②已知x\x²-4x-1=-1\3,求x四次方+2x+1\x五次。(我会给你们多加分的)
②已知x,y,z互不相等,且x+1\y=y+1\z=z+1\x,求证x²y²z²=1.
③已知x,y,z互不相等,且x+2\y=y+2\z=z+2\x=K,求K的值.
④已知x.y.z满足x+1\y=4,y+1\z=1,z+1\x=7\3,求xyz的值.
⑤已知a+b+c=0,且1\(a+1)+1\(b+2)+1\(c+3)=0,求(a+1)²+(b+2)²+(c+3)²的值.
sorry 再加两题①已知x²-4x+1=0,求x²\x四次方+x²+1.②已知x\x²-4x-1=-1\3,求x四次方+2x+1\x五次。(我会给你们多加分的)
1、a/b=b/c=c/d=d/a=k
则:a=bk,b=ck,c=dk,d=ak
a=bk=ck^2=dk^3=ak^4
所以k^4=1
如果a,b,c,d都是实数,那么k=+-1
所以得
a=b=c=d
或
a=-b=c=-d
代入要求的式子,得:
(a-b+c-d)/(a+b-c+d)=0或-2
2、令X+1/Y=Y+1/Z=Z+1/X=1
则 X=1-1/Y
=-(1-Y)/Y
Y=1-1/Z
则 Z=1/(1-Y)
(XYZ)^2=[-(1-Y)/Y*Y*1/(1-Y)]^2=(-1)^2=1
3、Y=2/(K-X) Z=K-2/X 又Y+2/Z=K 带入解得:(K^2-2)X^2+(2K-K^3)+2K^2-4=0又X,Y,Z各不相同又由题可看出X,Y,Z可互换,即可理解为X有唯一解,即(2K-K^3)^2-4(K^2-2)(2K^2-4)=0解得K^2=2或8,K就为正负根号2和正负2倍根号2
4、(X+1/Y)(Z+1/X)(Y+1/Z)=XYZ+1/(XYZ)+X+Y+Z+1/X+1/Y+1/Z=XYZ+1/(XYZ)+4+1+7/3=4*1*7/3
得到:
XYZ+1/(XYZ)=2,解得XYZ=1
5、a+b+c=0,所以(a+1)+(b+2)+(c+3)=6,
1/a+1+1/b+2+1/c+3=0,得(b+2)(a+1)+(b+2)(c+3)+(a+1)(c+3)=0
那么,(a+1)的平方+(b+2)的平方+(c+3)的平方
=〔(a+1)+(b+2)+(c+3)〕^2-2〔(b+2)(a+1)+(b+2)(c+3)+(a+1)(c+3)〕
=36-0
=36
则:a=bk,b=ck,c=dk,d=ak
a=bk=ck^2=dk^3=ak^4
所以k^4=1
如果a,b,c,d都是实数,那么k=+-1
所以得
a=b=c=d
或
a=-b=c=-d
代入要求的式子,得:
(a-b+c-d)/(a+b-c+d)=0或-2
2、令X+1/Y=Y+1/Z=Z+1/X=1
则 X=1-1/Y
=-(1-Y)/Y
Y=1-1/Z
则 Z=1/(1-Y)
(XYZ)^2=[-(1-Y)/Y*Y*1/(1-Y)]^2=(-1)^2=1
3、Y=2/(K-X) Z=K-2/X 又Y+2/Z=K 带入解得:(K^2-2)X^2+(2K-K^3)+2K^2-4=0又X,Y,Z各不相同又由题可看出X,Y,Z可互换,即可理解为X有唯一解,即(2K-K^3)^2-4(K^2-2)(2K^2-4)=0解得K^2=2或8,K就为正负根号2和正负2倍根号2
4、(X+1/Y)(Z+1/X)(Y+1/Z)=XYZ+1/(XYZ)+X+Y+Z+1/X+1/Y+1/Z=XYZ+1/(XYZ)+4+1+7/3=4*1*7/3
得到:
XYZ+1/(XYZ)=2,解得XYZ=1
5、a+b+c=0,所以(a+1)+(b+2)+(c+3)=6,
1/a+1+1/b+2+1/c+3=0,得(b+2)(a+1)+(b+2)(c+3)+(a+1)(c+3)=0
那么,(a+1)的平方+(b+2)的平方+(c+3)的平方
=〔(a+1)+(b+2)+(c+3)〕^2-2〔(b+2)(a+1)+(b+2)(c+3)+(a+1)(c+3)〕
=36-0
=36
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