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\(\hept{\begin{cases}xyz=12\\x^3+y^3+z^3=36\end{cases}}\Leftrightarrow x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)-3xyz+z^3=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
\(\Leftrightarrow x=y=z\left(x+y+z>0\right)\)
Thay x=y=z vào r tính thôi bạn
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
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
Bạn thay y xyz=2010 vào A ta được
A= xyz*x/xy+xyz*x+xyz + y/yz+y+xyz + z/xz+z+1
suy ra A=x^2yz/xy(1+xz+z) + y/y(z+1+xz) + z/xz+x+1
A= xz/1+xz+z + 1/z+1+xz + x/xz+z+1 = xz+1+x/xz+1+x =1
Vay A=1
Thay xyz = 2011 vào N được :
\(N=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}=\frac{xy.xz}{xy\left(z+xz+1\right)}+\frac{y}{y\left(z+xz+1\right)}+\frac{z}{z+xz+1}\)
\(=\frac{xz}{z+xz+1}+\frac{1}{z+xz+1}+\frac{z}{z+xz+1}=\frac{z+xz+1}{z+xz+1}=1\)
\(A=\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)
\(A=\dfrac{1}{xy+x+xyz}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)
\(A=\dfrac{1}{x\left(y+1+yz\right)}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)
\(A=\dfrac{xyz}{x\left(y+1+yz\right)}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)
\(A=\dfrac{yz}{y+1+yz}+\dfrac{1}{y+yz+1}+\dfrac{1}{xz+z+1}\)
\(A=\dfrac{yz+1}{y+1+yz}+\dfrac{1}{xz+z+1}\)
\(A=\dfrac{yz+xyz}{y+xyz+yz}+\dfrac{1}{xz+z+1}\)
\(A=\dfrac{y\left(z+xz\right)}{y\left(1+xz+z\right)}+\dfrac{1}{xz+z+1}\)
\(A=\dfrac{z+xz+1}{xz+z+1}\)
\(A=1\)