Question

Giải bài toán sau:

Answer 1

\(\left\{{}\begin{matrix}a+b+c=3\\a^2+b^2+c^2=3\end{matrix}\right.\)

=> (a+b+c)²= 9

=> ab + bc + ca =3 

=>\(a^2+b^2+c^2=ab+ac+cb\)

=>\(2a^2+2b^2+2c^2=2ab+2bc+2ca\)

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

=>a-b=0, b-c=0, c-a=0 => a=b=c

mà a+b+c=3 suy ra a=b=c=1 

Vậy A=\(\dfrac{2^2.2^2.2^2}{4.4.4}=1\)