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13 tháng 4 2020

x=1;y=-1;z=2 nhé bn đấy là tìm mò còn lời giải để mình nghĩ cái ( hơi lâu đấy =((( )

2 tháng 5 2016

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)

7 tháng 5 2021

help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

7 tháng 5 2021

hộ caiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

\(A+B+C=x^2yz+xy^2z+xyz^2=xyz\left(x+y+z\right)=xyz\)

6 tháng 3 2022

\(A=x^2yz\) \(B=xy^2z\) \(C=xyz^2\)

\(A+B+C=x^2yz+xy^2z+xyz^2\)

                    \(=xyz\left(x+y+z\right)=xyz.1=xyz\)