Ta có \(\frac{x+2xy+1}{x+xy+xz+1}=\frac{x+2xy+xyz}{x+xy+xz+xyz}=\frac{1+2y+yz}{\left(y+1\right)\left(z+1\right)}\)

Tương tự => \(M=\frac{1+2y+yz}{\left(y+1\right)\left(z+1\right)}+\frac{1+2z+zx}{\left(1+x\right)\left(z+1\right)}+\frac{1+2x+xy}{\left(1+x\right)\left(y+1\right)}\)

=> \(M=\frac{\left(1+2y+yz\right)\left(1+x\right)+\left(1+2z+zx\right)\left(1+y\right)+\left(1+2x+xy\right)\left(1+z\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)

=>\(M=\frac{6+3\left(x+y+z\right)+3\left(xy+yz+xz\right)}{2+\left(x+y+z\right)+\left(xy+yz+xz\right)}=3\)

\(A=\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{1}{xy+x+xyz}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{1}{x\left(y+1+yz\right)}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{xyz}{x\left(y+1+yz\right)}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{yz}{y+1+yz}+\dfrac{1}{y+yz+1}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{yz+1}{y+1+yz}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{yz+xyz}{y+xyz+yz}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{y\left(z+xz\right)}{y\left(1+xz+z\right)}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{z+xz+1}{xz+z+1}\)

\(A=1\)

\(A=\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)⇔\(A=\dfrac{z}{1+xz+z}+\dfrac{xz}{z+1+xz}+\dfrac{1}{xz+z+1}\)(vì xyz=1)

⇔\(A=\dfrac{z+xz+1}{xz+z+1}\)⇔\(A=1\)

Xong rồi nè bn ơi 

\(\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

\(=\dfrac{1}{\dfrac{1}{z}+\dfrac{1}{yz}+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{\dfrac{1}{y}+z+1}\)

\(=\dfrac{1}{\dfrac{y+1+yz}{yz}}+\dfrac{1}{yz+y+1}+\dfrac{1}{\dfrac{1+zy+y}{y}}\)

\(=\dfrac{yz}{y+1+yz}+\dfrac{1}{yz+y+1}+\dfrac{y}{1+zy+y}=\dfrac{y+yz+1}{y+yz+1}=1\)

ta có x/xy+x+1 +y/yz+y+1 +z/xz+z+1

=xz/xyz+xz+z +xyz/xyz^2+xyz+xz +z/xz+z+1

=xz/1+xz+z +1/z+1+xz +z/ xz+z+1

=xz+z+1 /xz+z+1 =1

giúp mình với mọi người ơi

a, A=xy+7x-3y-21                                                         b,B= xyz+xz-yz-z+xy+x-y-1

    A=(xy+7x)-(3y+21)                                                      B=(xyz+xz)-(yz+z)+(xy+x)-(y+1)

    A=x(y+7)-3(y+7)                                                          B=xz(y+1)-z(y+1)+x(y+1)-(y+1)

    A=(y+7)(x-3)                                                                B=(y+1)(xz-z+x-1)

Thay x=103, y=-17 vào biểu thức ta có:                         B=(y+1)[(xz-z)+(x-1)]

A=(-17+7)(103-3)                                                            B=(y+1)[z(x-1)+(x-1)]

A=(-10)(100)                                                                   B=(y+1)(x-1)(z+1)

A=-1000                                                                          Thay x=-9, y=-21, z=-31 vào biểu thức ta có:

                                                                                           B=(-21+1)(-9-1)(-31+1)

                                                                                           B=(-20)(-10)(-30)

                                                                                           B=200(-30)

                                                                                           B=-6000

a,A=xy+7x-3y-21

     =(xy-3y)+(7x-21)

     =y(x-3)+7(x-3)

     =(x-3)(y+7)

Thay x=103 và y=-17 vào biểu thức trên ta có:

A=(103-3)(-17+7)=100.(-10)=-1000

b,B=xyz+xz-yz-z+xy+x-y-1

     =(xyz+xz)-(yz+z)+(xy+x)-(y+1)

     =xz(y+1)-z(y+1)+x(y+1)-(y+1)

     =(y+1)(xz-z+x-1)

     =(y+1)[(xz-z)+(x-1)]

     =(y+1)[z(x-1)+(x-1)]

     =(y+1)(x-1)(z+1)

Thay x=-9 ;y=-21 và z=-31 vào biểu thức trên ta có:

B=(-21+1)(-9-1)(-31+1)=-20.(-10).(-30)=-6000

\(A=\frac{x}{xy+x+1}+\frac{y}{y+1+yz}+\frac{z}{1+z+xz}\)

\(=\frac{x}{xy+x+1}+\frac{xy}{xy+x+xyz}+\frac{xyz}{xy+xyz+x^2yz}\)

\(=\frac{x}{xy+x+1}+\frac{xy}{xy+x+1}+\frac{1}{xy+1+x}\)

\(=\frac{xy+x+1}{xy+x+1}=1\)

\(\frac{x}{xy+x+1}+\frac{xy}{yx+x+xyz}+\frac{xyz}{xy+xyz+x^2yz}\)

\(\frac{x}{xy+x+1}+\frac{xy}{yx+x+1}+\frac{1}{xy+1+x}\)

\(\frac{x+xy+1}{xy+x+1}=1\)

\(A=\dfrac{xyz.x}{xy+xyz.x+xyz}+\dfrac{y}{yz+y+2019}+\dfrac{yz}{xyz+yz+y}\)

\(=\dfrac{xz}{1+xz+z}+\dfrac{y}{yz+y+2019}+\dfrac{yz}{yz+y+2019}\)

\(=\dfrac{xyz}{y+xyz+yz}+\dfrac{y}{yz+y+2019}+\dfrac{yz}{yz+y+2019}\)

\(=\dfrac{2019}{y+2019+yz}+\dfrac{y}{yz+y+2019}+\dfrac{yz}{yz+y+2019}\)

\(=\dfrac{yz+y+2019}{yz+y+2019}=1\)

\(P=\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+xz}.\)

\(P=\frac{1}{1+x+xy}+\frac{x}{x\left(1+y+yz\right)}+\frac{xy}{xy\left(1+z+xz\right)}\)

\(P=\frac{1}{1+x+xy}+\frac{x}{x+xy+xyz}+\frac{xy}{xy+xyz+x^2yz}\)

\(P=\frac{1}{1+x+xy}+\frac{x}{x+xy+xyz}+\frac{xy}{xy+xyz+xyz.x}\)

\(P=\frac{1}{1+x+xy}+\frac{x}{x+xy+1}+\frac{xy}{xy+1+x}\left(xyz=1\right)\)

\(P=\frac{1+x+xy}{1+x+xy}=1\)

Vậy P=1

Hello hikaru nakamura

k ai trả lời đc ah