\(4x=7y\Rightarrow\dfrac{x}{7}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{49}=\dfrac{y^2}{16}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x^2}{49}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{49+16}=\dfrac{260}{65}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=196\\y^2=64\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-14;y=-8\\x=14;y=8\end{matrix}\right.\)
$4a=7b\Leftrightarrow \dfrac{a}{7}=\dfrac{b}{4}$
$\Leftrightarrow \dfrac{a^2}{49}=\dfrac{b^2}{16}$
Áp dụng tính chất dãy tỉ số bằng nhau:
$\dfrac{a^2}{49}=\dfrac{b^2}{16}=\dfrac{a^2+b^2}{49+16}=\dfrac{260}{65}=4$
$\Rightarrow \begin{cases}\dfrac{a^2}{49}=4\\\dfrac{b^2}{16}=4\end{cases}$
$\Leftrightarrow\begin{cases}a^2=196\\b^2=64\end{cases}$
$\Leftrightarrow \begin{cases}a=\pm 14\\b=\pm 8\end{cases}$
Vậy $a=\pm 14;b=\pm 8$
Ta có: 4a=7b
nên \(\dfrac{a}{\dfrac{1}{4}}=\dfrac{b}{\dfrac{1}{7}}\)
Đặt \(\dfrac{a}{\dfrac{1}{4}}=\dfrac{b}{\dfrac{1}{7}}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{4}k\\b=\dfrac{1}{7}k\end{matrix}\right.\)
Ta có: \(a^2+b^2=260\)
\(\Leftrightarrow k^2\cdot\dfrac{1}{16}+k^2\cdot\dfrac{1}{49}=260\)
\(\Leftrightarrow k^2=3136\)
Trường hợp 1: k=56
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{4}k=\dfrac{1}{4}\cdot56=14\\b=\dfrac{1}{7}k=\dfrac{1}{7}\cdot56=8\end{matrix}\right.\)
Trường hợp 2: k=-56
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{4}k=\dfrac{1}{4}\cdot\left(-56\right)=-14\\b=\dfrac{1}{7}k=\dfrac{1}{7}\cdot\left(-56\right)=-8\end{matrix}\right.\)
⇔a249=b216⇔a249=b216
Áp dụng tính chất dãy tỉ số bằng nhau:
⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩a249=4b216=4⇒{a249=4b216=4
⇔{a2=196b2=64⇔{a2=196b2=64
⇔{a=±14b=±8⇔{a=±14b=±8
Vậy a=±14;b=±8
