\(\frac{a}{b}=\frac{2}{3}\Rightarrow\frac{a}{3}=\frac{b}{3}\)
\(\Rightarrow\frac{a^2}{4}=\frac{b^2}{9}=\frac{a^2+b^2}{4+9}=\frac{208}{13}=16\)
\(\Rightarrow a^2=4.16=64\Rightarrow a=8\) (vì \(a\in N\)*)
\(b^2=9.16=144\Rightarrow b=12\) (vì \(b\in N\)*)
Giải:
Ta có: \(\frac{a}{b}=\frac{2}{3}\Rightarrow\frac{a}{2}=\frac{b}{3}\)
Đặt \(\frac{a}{2}=\frac{b}{3}=k\Rightarrow a=2k,b=3k\)
Mà \(a^2+b^2=208\)
\(\Rightarrow\left(2k\right)^2+\left(3k\right)^2=208\)
\(\Rightarrow2^2.k^2+3^2.k^2=208\)
\(\Rightarrow k^2.\left(2^2+3^2\right)=208\)
\(\Rightarrow k^2.13=208\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k=\pm4\)
Mà \(a,b\in\) N*
\(\Rightarrow k=4\)
\(\Rightarrow a=8,b=12\)
Vậy \(a=8,b=12\)
