Question

Cho ba số a, b, c khác 0 thỏa mãn \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\). Hãy tính P = \(\dfrac{ac}{c^2}+\dfrac{bc}{a^2}+\dfrac{ac}{b^2}\)

Answer 1

Đặt x = \(\dfrac{1}{a}\); y = \(\dfrac{1}{b}\); z = \(\dfrac{1}{c}\); x + y + z = 0 (vì \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\))

x = -(y + z)

x³ + y³ + z³ - 3xyz = - (y + z)³ + y³ + z³ - 3xyz

-(y³ + 3y²z + 3y²z² + z³) + y³ + z³ - 3xyz = -3yz(y + z + x) = -3yz . 0 = 0

Từ x³ + y³ +z³ - 3xyz = 0 \(\Leftrightarrow\) x³ + y³ +z³ = 3xyz

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

Nếu \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) thì \(\)P = \(\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ac}{b^2}=3\)