Question

A) CHO \(ABC\ne0\)VÀ \(A+B+C=\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\).CM RẰNG \(B\left(A^2-BC\right)\left(1-AC\right)=A\left(1-BC\right)\left(B^2-AC\right)\)

Answer 1

Ta có:

\(a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

\(\Leftrightarrow abc^2+ab^2c+a^2bc-ab-bc-ca=0\left(1\right)\)

Ta cần chứng minh

\(b\left(a^2-bc\right)\left(1-ac\right)=a\left(1-bc\right)\left(b^2-ac\right)\)

\(\Leftrightarrow ab^2c^2-a^2bc^2+ab^3c-b^2c-a^3bc+a^2c-ab^2+a^2b=0\)

\(\Leftrightarrow b\left(abc^2+ab^2c-bc-ab\right)-a^2bc^2-a^3bc+a^2c+a^2b=0\)

\(\Leftrightarrow b\left(ac-a^2bc\right)-a^2bc^2-a^3bc+a^2c+a^2b=0\)

\(\Leftrightarrow-a\left(ab^2c+abc^2+a^2bc-bc-ac-ab\right)=0\)(theo (1) thì đúng)

\(\RightarrowĐPCM\)