Question

[tèn tén ten]

Tìm a, b thuộc N* biết a/b = 2/3; a^2 + b^2 =208

Answer 1

[Nguyễn Anh Duy]

\(\frac{a}{b}=\frac{2}{3}\Rightarrow\frac{a}{3}=\frac{b}{3}\)

\(\Rightarrow\frac{a^2}{4}=\frac{b^2}{9}=\frac{a^2+b^2}{4+9}=\frac{208}{13}=16\)

\(\Rightarrow a^2=4.16=64\Rightarrow a=8\) (vì \(a\in N\)*)

\(b^2=9.16=144\Rightarrow b=12\) (vì \(b\in N\)*)

Answer 2

[Nguyễn Huy Tú]

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 \(a=8,b=12\)