Câu hỏi của nguyen kim chi

Question

cho 3 so huu ti a;b;c thoa man 1/a+1/b=1/c

c/m $\sqrt{a^2+b^2+c^2}$ la so huu ti

Mr Lazy · Accepted Answer

$\frac{1}{a}+\frac{1}{b}=\frac{1}{c}\Rightarrow c=\frac{ab}{a+b}$

$a^2+b^2+c^2=\left(a+b\right)^2-2ab+\frac{a^2b^2}{\left(a+b\right)^2}=\frac{\left(a+b\right)^4-2ab\left(a+b\right)^2+a^2b^2}{\left(a+b\right)^2}$

$=\frac{\left[\left(a+b\right)^2-ab\right]^2}{\left(a+b\right)^2}$

$\Rightarrow\sqrt{a^2+b^2+c^2}=\left|\frac{\left(a+b\right)^2-ab}{a+b}\right|$ là số hữu tỉ.

Câu trả lời:

$\frac{1}{a}+\frac{1}{b}=\frac{1}{c}\Rightarrow c=\frac{ab}{a+b}$

$a^2+b^2+c^2=\left(a+b\right)^2-2ab+\frac{a^2b^2}{\left(a+b\right)^2}=\frac{\left(a+b\right)^4-2ab\left(a+b\right)^2+a^2b^2}{\left(a+b\right)^2}$

$=\frac{\left[\left(a+b\right)^2-ab\right]^2}{\left(a+b\right)^2}$

$\Rightarrow\sqrt{a^2+b^2+c^2}=\left|\frac{\left(a+b\right)^2-ab}{a+b}\right|$ là số hữu tỉ.