Lời giải:
Ta có \(P=\frac{a^2}{(a-b)(a-c)}+\frac{b^2}{(b-c)(b-a)}+\frac{c^2}{(c-a)(c-b)}\)
\(=\frac{a^2(c-b)}{(a-b)(b-c)(c-a)}+\frac{b^2(a-c)}{(a-b)(b-c)(c-a)}+\frac{c^2(b-a)}{(a-b)(b-c)(c-a)}\)
\(=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(a-b)(b-c)(c-a)}\)
Thực hiện khai triển suy ra:\((a-b)(b-c)(c-a)=(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)\)
\(\Rightarrow P=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}=1\)
\(\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^2}{\left(b-c\right)\left(b-a\right)}+\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}-\dfrac{b^2}{\left(b-c\right)\left(a-b\right)}-\dfrac{c^2}{\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{a^2b-a^2c-ab^2+b^2c+ac^2-bc^2}{a^2b-a^2c-abc+ac^2-ab^2+abc+b^2c-bc^2}\)
\(=\dfrac{a^2b-a^2c-ab^2+b^2c+ac^2-bc^2}{a^2b-a^2c-ab^2+b^2c+ac^2-bc^2}\)
\(=1\)
Tick cho mk nha