Question

Tính

\(\dfrac{x^2}{\left(x-y\right)\left(x-z\right)}+\dfrac{y^2}{\left(y-z\right)\left(y-x\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)}\)

Answer 1

Lời giải:

Đặt biểu thức cần tính là A

Ta có:

\(A=\frac{x^2}{(x-y)(x-z)}+\frac{y^2}{(y-z)(y-x)}+\frac{z^2}{(z-x)(z-y)}\)

\(A=\frac{-x^2}{(x-y)(z-x)}+\frac{-y^2}{(y-z)(x-y)}+\frac{-z^2}{(z-x)(y-z)}\)

\(A=\frac{-x^2(y-z)+(-y^2)(z-x)+(-z)^2(x-y)}{(x-y)(y-z)(z-x)}\)

\(\text{ tử số}=x^2(z-y)+y^2(x-z)+z^2(y-x)\)

\(=x^2(z-y)-y^2[(z-y)+(y-x)]+z^2(y-x)\)

\(=(z-y)(x^2-y^2)+(y-x)(z^2-y^2)\)

\(=(z-y)(x-y)(x+y)-(x-y)(z-y)(z+y)\)

\(=(x-y)(z-y)(x+y-z-y)=(x-y)(z-y)(x-z)=(x-y)(y-z)(z-x)\)

Do đó: \(A=\frac{(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=1\)