cho x+y+z=0,xy+yz+zx=1 tinh p=x^2(1+y^2).(1+z^2)/(1+x^2) + y^2(1+z^2)(1+x^2)/(1+y^2) + z^2(1+x^2)(1+y^2)/1+z^2

0 bình luận về “cho x+y+z=0,xy+yz+zx=1 tinh p=x^2(1+y^2).(1+z^2)/(1+x^2) + y^2(1+z^2)(1+x^2)/(1+y^2) + z^2(1+x^2)(1+y^2)/1+z^2”

1. Đáp án:

    

   Giải thích các bước giải:

   `P=[x^2(1+y^2).(1+z^2)]/(1+x^2) + [y^2(1+z^2)(1+x^2)]/(1+y^2) + [z^2(1+x^2)(1+y^2)]/(1+z^2)`

   `+)[x^2(1+y^2).(1+z^2)]/(1+x^2)`

   `=[x^2(y^2+xy+yz+zx)(z^2+xy+yz+zx)]/(x^2+xy+yz+zx)`

   `=[x^2(x+y)(y+z)(x+z)(y+z)]/[(x+y)(x+z)]` `=x^2(y+z)^2`

   CMTT

   `[y^2(1+z^2)(1+x^2)]/(1+y^2)=y^2(z+x)^2` `[z^2(1+x^2)(1+y^2)]/(1+z^2)=z^2(x+y)^2`

   Cộng từng vế ta có `P=x^2(y+z)^2+y^2(z+x)^2+z^2(x+y)^2`

   `x+y+z=0=>y+z=-x,x+y=-z,x+z=-y`

   `P=x^2(-x)+y^2(-y)+z^2(-z)`

   `=-x^3-y^3-z^3` `=-(x^3+y^3+z^3)`

   `=-[(x+y)^3+z^3-3xy(x+y)]`

   `=-[(x+y+z)(x^2+y^2+z^2+2xy-xz-yz)-3xy(-z)]`

   `=-[0+3xyz]` `=-3xyz`

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