Question

Cho a+b+c=1, \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) abc khác 0.Tính A = a² + b² + c²

Answer 1

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc}{abc}+\frac{ac}{abc}+\frac{ab}{abc}\)

\(\Rightarrow\frac{bc+ac+ab}{abc}=0\)

\(\Rightarrow bc+ca+ab=0\)

\(\Rightarrow2bc+2ac+2ab=0\)

Đặt \(B=a^2+b^2+c^2+2ab+2bc+2ca\)

\(\Rightarrow B=\left(a+b+c\right)^2=1^2=1\) ( áp dụng hằng đẳng thức )

\(\Rightarrow B=a^2+b^2+c^2+0=1\)

\(\Rightarrow A=a^2+b^2+c^2=1-0=1\)

Vậy \(A=1\)

Answer 2

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\frac{bc+ac+ab}{abc}=0\)

=> ac + ab + bc = 0

a² + b² + c² 

= (a + b + c)² - 2(ab + ac + bc)

= 1² - 2 . 0

= 1