Question

[Postgass D Ace]

Cho \(P=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)và \(Q=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)với \(a\ne b\ne c\),    a,b,c khác 0 và \(a^3+b^3+c^3=3abc\) CMR: PQ=9

Answer 1

[Thanh Tùng DZ]

Chứng minh : a³ + b³ + c³ = 3abc \(\Rightarrow\orbr{\begin{cases}a+b+c=0\left(tm\right)\\a=b=c\left(loai\right)\end{cases}}\)

Rút gọn P

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

Xét : ab(a-b) + bc(b-c) + ac(c-a) = ab[-(b-c)-(c-a)] + bc(b-c) + ac(c-a)

= (b-c)(bc-ab) + (c-a)(ac-ab) = b(b-c)(c-a) + a(c-a)(c-b) = (c-a)(c-b)(a-b)

\(\Rightarrow P=\frac{\left(c-a\right)\left(c-b\right)\left(a-b\right)}{abc}\)

Rút gọn Q

Đặt a - b = z ; b-c = x ; c - a = y

\(\Rightarrow\)x- y = a + b - 2c = -c - 2c = -3c              ( do a + b + c = 0 )

y - z = -3a ; z - x = -3b

\(\Rightarrow\)\(-3Q=\frac{\left(y-z\right)}{x}+\frac{\left(z-x\right)}{y}+\frac{\left(x-y\right)}{z}\)

Làm tương tự như rút gọn P, ta có : 

\(-3Q=\frac{\left(x-y\right)\left(z-y\right)\left(z-x\right)}{xyz}=\frac{-\left(-3a\right)\left(-3b\right)\left(-3c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{27abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{-27abc}{\left(a-b\right)\left(c-b\right)\left(c-a\right)}\)

\(\Rightarrow Q=\frac{9abc}{\left(a-b\right)\left(c-b\right)\left(c-a\right)}\)

\(\Rightarrow PQ=9\)