Question

Cho \(a+b+c=a^2+b^2+c^2=1\) và \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) \(\left(a\ne0,b\ne0,c\ne0\right)\)

Chứng minh rằng: \(\left(x+y+z\right)^2=x^2+y^2+z^2\)

Answer 1

Lời giải:

Đặt $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=t$

$\Rightarrow x=at; y=bt; z=ct$. Ta có:

$(x+y+z)^2=(at+bt+ct)^2=t^2(a+b+c)^2=t^2(*)$

Mặt khác:

$x^2+y^2+z^2=(at)^2+(bt)^2+(ct)^2=t^2(a^2+b^2+c^2)=t^2(**)$

Từ $(*); (**)\Rightarrow (x+y+z)^2=x^2+y^2+z^2$ (đpcm)