Question

[Thùy Dung A]

Cho a, b, c là 3 số đôi một khác nhau và khác 0. Thỏa mãn a³ + b³ + c³ =3abc

Tính giá trị biểui thức: P = \(\frac{a^2}{bc}\)+ \(\frac{b^2}{ac}\)+ \(\frac{c^2}{ab}\)

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

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

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Answer 1

[Đinh Đức Hùng]

Từng ý nhé !!!

\(P=\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=\frac{1}{abc}\left(a^3+b^3+c^3\right)\)

\(\frac{1}{abc}.3abc=3\)

Answer 2

[Đinh Đức Hùng]

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

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

\(\Leftrightarrow\left(a+b+c\right)\left[\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)

Xét \(a+b+c=0\) ta có :\(\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)

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

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

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

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

\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{1}{2abc}\left(a^3+b^3+c^3\right)=\frac{1}{2abc}.3abc=\frac{3}{2}\)

Xét \(a=b=c\) ta có :

\(Q=\frac{a^2}{a^2-a^2-a^2}+\frac{b^2}{b^2-b^2-b^2}+\frac{c^2}{c^2-c^2-c^2}=-1-1-1=-3\)