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![](https://rs.olm.vn/images/avt/0.png?1311)
`A + B + C = x^2yz + xy^2z + zy^2x = xyz(x+y+z) = xyz`.
\(A+B+C=x^2yz+xy^2z+xyz^2=xyz\left(x+y+z\right)=xyz\)
Vậy ta có đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(A+B+C\)
\(=x^2yz+xy^2z+xyz^2\)
\(=xyz\left(x+y+z\right)\)
\(=xyz.1\)
\(=xyz\left(đpcm\right)\)
ta có: A + B + C = x2yz +xy2z + xyz2 = xyz.(x+y+z) = xyz.1=xyz
=> A+B+C = xyz
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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 4:
b: \(=x^2z\left(-1+3-7\right)=-5x^2z=-5\cdot\left(-1\right)^2\cdot\left(-2\right)=10\)
c: \(=xy^2\left(5+0.5-3\right)=2.5xy^2=2.5\cdot2\cdot1^2=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
\(A\left(x\right)+B\left(x\right)\)
\(=\left(6x-4x^3+x-1\right)+\left(-3x-2x^3-5x^2+x+2\right)\)
\(=\left(6x+-3x+x\right)-\left(4x^3+2x^3\right)-5x^2+\left(-1+2\right)\)
\(=-6x^3-5x^2+4x+1\)
\(A\left(x\right)-B\left(x\right)\)
\(=\left(6x-4x^3+x-1\right)-\left(-3x-2x^3-5x^2+x+2\right)\)
\(=\left(-4x^3+2x^3\right)+5x^2+\left(6x+x-x\right)+\left(-1-2\right)\)
\(=-2x^3+5x^2+6x-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A+B+C=x^2yz+xy^2z+xyz^2=xyz\left(x+y+z\right)=xyz\)
\(A=x^2yz\) \(B=xy^2z\) \(C=xyz^2\)
\(A+B+C=x^2yz+xy^2z+xyz^2\)
\(=xyz\left(x+y+z\right)=xyz.1=xyz\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Ta\) \(có:\)
\(A+B+C=x^2yz+xy^2z+xyz^2=xyz\left(x+y+z\right)=xyz.1=xyz\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta cs: \(A+B+C=x^2yz+xy^2z+xyz^2\)
\(=xyz\left(x+y+z\right)\)
Mà x+y+z=1
=>A+B+C=xyz.1=xyz(đpcm)