Question

cho a,b,c đôi một khác nhau tính giá trị của biểu thức 

P=\(\frac{a^2}{\left(a-b\right)\left(a-c\right)}\)+\(\frac{b^2}{\left(b-a\right)\left(b-c\right)}\)+\(\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)

Answer 1

\(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)

\(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(a-b\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)

\(P=\frac{1}{a-b}.\frac{a^2\left(b-c\right)-b^2\left(a-c\right)}{\left(a-c\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)

\(P=\frac{1}{a-b}.\frac{a^2b-a^2c-b^2a+b^2c}{\left(a-c\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)

\(P=\frac{1}{a-b}.\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)}{\left(a-c\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)

\(P=\frac{1}{a-b}.\frac{\left(a-b\right)\left(ab-ac-bc\right)}{\left(a-c\right)\left(b-c\right)}+\frac{c^2}{\left(b-c\right)\left(a-c\right)}\)

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

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

=> P = 1

Đáp số: P=1

Answer 2

\(P=-\frac{a^2}{\left(a-b\right)\left(c-a\right)}-\frac{b^2}{\left(a-b\right)\left(b-c\right)}-\frac{c^2}{\left(c-a\right)\left(b-c\right)}\)

\(=-\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=-\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)