Question

cho các số a, b , c khác 0 thỏa mãn 1/a + 1/b + 1/c = 2

và 1/a^2 + 1/b^2 + 1/ c^2 = 2

tính biểu thức P = a + b + c - abc

các bạn giúp mk vs nhé

Answer 1

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2=>\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)

\(=>\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=4\)

Thay \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\)

\(2+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=4=>\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=1\)

=>\(\dfrac{a+b+c}{abc}=1\)

=>a+b+c=abc => a+b+c-abc=0