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

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

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

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

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

\(=1\)

đpcm

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

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)

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 = \(\frac{xz+z+1}{xz+z+1}=1\)

Thay xyz = 2011 vào N được : 

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

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

\(1.\) Giả sử : \(a\ge b\ge c\Rightarrow a+b\ge a+c\ge b+c\)

Ta có : \(\dfrac{c}{a+b}\le\dfrac{c}{b+c};\dfrac{b}{a+c}\le\dfrac{b}{b+c};\dfrac{a}{b+c}=\dfrac{a}{b+c}\)

\(\Rightarrow\dfrac{c}{a+b}+\dfrac{b}{a+c}+\dfrac{a}{b+c}\le\dfrac{b+c}{b+c}+\dfrac{a}{b+c}=1+\dfrac{a}{b+c}< 1+1=2\left(đpcm\right)\)

\(2.\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)

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

\(\Leftrightarrow\left(x+y+z\right)\left(xy+yz+xz\right)=xyz\)

\(\Leftrightarrow x^2y+x^2z+xy^2+y^2z+xyz+xyz+yz^2+xz^2=0\)

\(\Leftrightarrow xy\left(x+y+z\right)+yz\left(x+y+z\right)+xz\left(x+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)y\left(x+z\right)+xz\left(x+z\right)=0\)

\(\Leftrightarrow\left(x+z\right)\left(xy+y^2+yz+xz\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=-z\\x=-z\end{matrix}\right.\)

+) Với : \(x=-y\) , ta có :

Đpcm \(\Leftrightarrow-\dfrac{1}{y^{2011}}+\dfrac{1}{y^{2011}}+\dfrac{1}{z^{2011}}=\dfrac{1}{-y^{2011}+y^{2011}+z^{2011}}\)

\(\Leftrightarrow\dfrac{1}{z^{2011}}=\dfrac{1}{z^{2011}}\left(luôn-đúng\right)\)

Tương tự với 2 TH còn lại .

\(\RightarrowĐCPM\)