Đặt \(\dfrac{1}{a}=\dfrac{1}{x+y},\dfrac{1}{b}=\dfrac{1}{y+z},\dfrac{1}{c}=\dfrac{1}{z+x}\)

Đề trở thành: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\), tính \(P=\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) Tương đương \(ab+bc=-ac\)

\(P=\dfrac{b^3c^3+a^3c^3+a^3b^3}{a^2b^2c^2}=\dfrac{\left(ab+bc\right)\left(a^2b^2-ab^2c+b^2c^2\right)+a^3c^3}{a^2b^2c^2}=\dfrac{-ac\left(a^2b^2-ab^2c+b^2c^2\right)+a^3c^3}{a^2b^2c^2}\)

\(=\dfrac{a^2c^2-a^2b^2+ab^2c-b^2c^2}{ab^2c}=\dfrac{ac}{b^2}-\dfrac{a}{c}+1-\dfrac{c}{a}\)\(=ac\left(\dfrac{1}{a^2}+\dfrac{2}{ac}+\dfrac{1}{c^2}\right)-\dfrac{a}{c}+1-\dfrac{c}{a}\) (do \(\dfrac{1}{b}=-\dfrac{1}{a}-\dfrac{1}{c}\) tương đương \(\dfrac{1}{b^2}=\dfrac{1}{a^2}+\dfrac{2}{ac}+\dfrac{1}{c^2}\)) 

\(=3\)

Vậy P=3

Hello

Sửa đề cho x/y-z + y/z-x + z/x-y =0,tính Q=x/(y-z)^2 + y/(z-x)^2 + z/(x-y)^2

Ta có: \(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y}=0\Rightarrow\frac{x}{y-z}=-\left(\frac{y}{z-x}+\frac{z}{x-y}\right)\)

\(\Rightarrow\frac{x}{y-z}=\frac{y}{x-z}+\frac{z}{y-x}=\frac{y^2-xy+xz-z^2}{\left(x-y\right)\left(z-x\right)}\)

\(\Rightarrow\frac{x}{\left(y-z\right)^2}=\frac{y^2-xy+xz-z^2}{\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)

Tương tự ta có: \(\frac{y}{\left(z-x\right)^2}=\frac{z^2-yz+yx-x^2}{\left(y-z\right)\left(z-x\right)\left(x-y\right)};\frac{z}{\left(x-y\right)^2}=\frac{x^2-zx+zy-y^2}{\left(z-x\right)\left(x-y\right)\left(y-z\right)}\)

Cộng ba đẳng thức trên vế theo vế, ta được:

\(\frac{x}{\left(y-z\right)^2}+\frac{y}{\left(z-x\right)^2}+\frac{z}{\left(x-y\right)^2}=\frac{y^2-xy+xz-z^2+z^2-yz+yx-x^2+x^2-zx+zy-y^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)

Vậy Q = 0

\(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1\Leftrightarrow\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)=x+y+z\)

<=>\(\frac{x^2+x\left(y+z\right)}{y+z}+\frac{y^2+y\left(z+x\right)}{z+x}+\frac{z^2+z\left(x+y\right)}{x+y}=x+y+z\)

<=>\(\frac{x^2}{y+z}+x+\frac{y^2}{z+x}+y+\frac{z^2}{x+y}+z=x+y+z\)

<=>\(S=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}=0\)

 x/(y+z)+y/(x+z)+z/(x+y)=1

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