Question

Cho x,y,z khác 0 và x+y+z=2008 . Tính giá trị biểu thức : \(P=\dfrac{x^3}{\left(x-y\right)\left(x-z\right)}+\dfrac{y^3}{\left(y-x\right)\left(y-z\right)}+\dfrac{z^3}{\left(z-y\right)\left(z-x\right)}\)

Answer 1

Lời giải:
Ta có:

\(P=\frac{x^3}{(x-y)(x-z)}+\frac{y^3}{(y-x)(y-z)}+\frac{z^3}{(z-y)(z-x)}\)

\(=\frac{x^3(y-z)+y^3(x-z)+z^3(y-x)}{(x-y)(y-z)(z-x)}\)

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

\(=\frac{xz(x-z)(x+z)+xy(y-x)(y+x)+zy(z-y)(z+y)}{(x-y)(y-z)(z-x)}\)

\(=\frac{xz(x-z)(2008-y)+xy(y-x)(2008-z)+zy(z-y)(2008-x)}{(x-y)(y-z)(z-x)}\)

\(=\frac{2008[xz(x-z)+xy(y-x)+zy(z-y)-xyz(x-z+y-x+z-y)}{(x-y)(y-z)(z-x)}\)

\(=\frac{2008[xz(x-z)+xy(y-x)+zy(z-y)]}{xz(x-z)+xy(y-x)+zy(z-y)}=2008\)