Thay xyz = 2011 vào N được : 

\(N=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}=\frac{xy.xz}{xy\left(z+xz+1\right)}+\frac{y}{y\left(z+xz+1\right)}+\frac{z}{z+xz+1}\)

\(=\frac{xz}{z+xz+1}+\frac{1}{z+xz+1}+\frac{z}{z+xz+1}=\frac{z+xz+1}{z+xz+1}=1\)

Phân thức thứ nhất

\(\frac{2011x}{xy+2011x+2011}=\frac{2011xz}{xyz+2011xz+2011z}=\frac{2011xz}{2011+2011xz+2011z}=\frac{2011xz}{2011\left(1+xz+z\right)}=\frac{xz}{xz+z+1}\)

Phân thức thứ hai

\(\frac{y}{yz+y+2011}=\frac{y}{yz+y+xyz}=\frac{y}{y\left(z+1+xz\right)}=\frac{1}{xz+z+1}\)

Cộng ba phân thức

=> biểu thức = \(\frac{xz+z+1}{xz+z+1}=1\)

\(\frac{2011x}{xy+2011x+2011}+\frac{y}{yz+y+2011}+\frac{z}{zx+z+1}\)

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

\(=\frac{x^2yz}{xy.\left(xz+z+1\right)}+\frac{y}{y.\left(xz+z+1\right)}+\frac{z}{zx+z+1}\)

\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{zx+z+1}\)

\(=\frac{xz+1+z}{xz+1+z}\)

\(=1\)

đpcm

Tại sao lại có nhìu đứa rảnh háng đi trả lời câu này nhỉ ?

b: 5x^2+5y^2+8xy-2x+2y+2=0

=>4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0

=>(x-1)^2+(y+1)^2+(2x+2y)^2=0

=>x=1 và y=-1

M=(1-1)^2015+(1-2)^2016+(-1+1)^2017=1

\(\dfrac{2011x}{xy+2011x+2011}+\dfrac{y}{yz+y+2011}+\dfrac{z}{xz+z+x}\)

\(=\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}\)

\(=\dfrac{x^2yz}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{xz+z+1}\)

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

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

\(=1\) ( Đpcm )

Bài này biến đổi cơ bản thế quái nào câu hỏi hay

Ta có: xyz = 2011

=>\(\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}\)

^{=> \(\dfrac{x^2yz}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{xz+z+1}\)}

^{=>\(\dfrac{xz}{xz+1+z}+\dfrac{1}{xz+1+z}+\dfrac{z}{xz+1+z}\)}

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

Do \(xyz=2011\Rightarrow\dfrac{xy}{2011}=\dfrac{1}{z}\)

\(\dfrac{2011x}{xy+2011x+2011}+\dfrac{y}{yz+y+2011}+\dfrac{z}{xz+z+1}\)

\(=\dfrac{x}{\dfrac{xy}{2100}+x+1}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}\)

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

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

\(=\dfrac{xz+z+1}{xz+z+1}=1\) (đpcm)

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

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

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