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31 tháng 3 2015

A + B + C = x2.y.z + x.y2.z + x.y.z2 = x.y.z.(x + y + z) = x.y.z .1 = xyz (Vì x+ y + z = 1)

21 tháng 11 2017

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

21 tháng 11 2017

\(A+B+C=xyz\)

\(VT=A+B+C\)

\(\Leftrightarrow VT=x^2yz+xy^2z+xyz^2\)

\(\Leftrightarrow VT=xyz\left(x+y+z\right)\)

\(\Leftrightarrow VT=xyz\)

\(\Rightarrow VT=VP\)

\(\Rightarrow A+B+C=xyz\left(dpcm\right)\)

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

6 tháng 3 2022

\(A=x^2yz\) \(B=xy^2z\) \(C=xyz^2\)

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

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

 

7 tháng 5 2021

help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

7 tháng 5 2021

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20 tháng 3 2016

Giúp mìk đj mìk K cho

ta có A+B+C=x2yz+xy2z+xyz2

=x(xyz)+y(xyz)+z(xyz)

=x.1+y.1+z.1

=x+y+z(dpcm)

18 tháng 4 2016

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

\(B=xy^2z=y.\left(xyz\right)=y.1=y\)

\(C=xyz^2=z.\left(xyz\right)=z.1=z\)

\(\Rightarrow A+B+C=x+y+z\)

29 tháng 4 2018

Ta có:

\(A+B+C=x^2yz+xy^2z+xyz^2\\ A+B+C=xyz\left(x+y+z\right)\\ A+B+C=xyz\times1\\ A+B+C=xyz\)

Vậy A+B+C=xyz

17 tháng 7 2023

Câu 1:

\(A\left(x\right)+B\left(x\right)\)

\(=\left(6x-4x^3+x-1\right)+\left(-3x-2x^3-5x^2+x+2\right)\)

\(=\left(6x+-3x+x\right)-\left(4x^3+2x^3\right)-5x^2+\left(-1+2\right)\)

\(=-6x^3-5x^2+4x+1\)

\(A\left(x\right)-B\left(x\right)\)

\(=\left(6x-4x^3+x-1\right)-\left(-3x-2x^3-5x^2+x+2\right)\)

\(=\left(-4x^3+2x^3\right)+5x^2+\left(6x+x-x\right)+\left(-1-2\right)\)

\(=-2x^3+5x^2+6x-3\)

20 tháng 4 2016
Xxyz+yxyz+zxyz=xyz(x+y+z)=xyz.1=xyz