Tìm x, y thỏa |5-3/4x|+|2/7y+3|=0

1. Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\) \(\left(1\right)\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) \(\left(2\right)\)
Từ \(\left(1\right)\text{và (2)}\) \(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}\)
2. \(\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0\)
\(\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|\ge0\\\left|\dfrac{2}{7}y+3\right|\ge0\end{matrix}\right.\Rightarrow\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|\ge0\)
\(\text{Mà }\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|=0\\\left|\dfrac{2}{7}y+3\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5-\dfrac{3}{4}x=0\\\dfrac{2}{7}y+3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{4}x=5\\\dfrac{2}{7}x=-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{20}{3}\\y=-\dfrac{21}{2}\end{matrix}\right.\)
\(\text{Vậy }\left\{{}\begin{matrix}x=\dfrac{20}{3}\\y=-\dfrac{21}{2}\end{matrix}\right.\)