Question

Cho :

                 \(a+b+c=a^2+b^2+c^2=1\)

         và 

                 \(x:y:z=a:b:c\)

Chứng minh rằng :

                        \(\left(x+y+z\right)^2=x^2+y^2+z^2\)

Answer 1

Ta có :

\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}=x+y+z\)( Vì a+b+c=1)

Do đó :

\(\left(x+y+z\right)^2=\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\)( Vì \(a^2+b^2+c^2=1\) ).

Vậy \(\left(x+y+z\right)^2=x^2+y^2+z^2.\)