Question

[tran khanh bac]

Cho: A=x^2yz;B=xy^2z;C=xyz^2 va x+y+z=1

hay chung to:A+B+C=xyz

Answer 1

[Hoàng Phúc]

Ta có:

\(A=x^2yz=x.x.y.z=x.xyz\left(1\right)\)

\(B=xy^2z=x.y.y.z=y.xyz\left(2\right)\)

\(C=xyz^2=x.y.z.z=z.xyz\left(3\right)\)

Lấy (1)+(2)+(3),vế theo vế ta được:

\(A+B+C=x.xyz+y.xyz+z.xyz=\left(x+y+z\right).xyz=xyz\) (vì x+y+z=1)

Vậy A+B+C=xyz      (đpcm)