\(a,x+y=\dfrac{x}{y}=3\left(x-y\right)\\ \text{Xét }x+y=3\left(x-y\right)\\ \Rightarrow x+y=3x-3y\\ \Rightarrow2x=-4y\Rightarrow x=-2y\\ \text{Thay vào }x+y=\dfrac{x}{y}\\ \Rightarrow-2y+y=\dfrac{-2y}{y}\\ \Rightarrow-3y=-2\Rightarrow y=\dfrac{2}{3}\\ \Rightarrow x=-\dfrac{4}{3}\\ \text{Vậy }\left(x;y\right)=\left(-\dfrac{4}{3};\dfrac{2}{3}\right)\)
\(b,A=\left|2016-2x\right|+\left|2x-2\right|-1999\\ \Rightarrow A\ge\left|2016-2x+2x-2\right|-1999=2014-1999=15\\ \Rightarrow A_{min}=15\Leftrightarrow\left(2016-2x\right)\left(2x-2\right)\ge0\\ \Rightarrow\left(1008-x\right)\left(x-1\right)\ge0\\ \Rightarrow1\le x\le1008\\ c,\dfrac{\overline{ab}}{\overline{bc}}=\dfrac{10a+b}{10b+c}=\dfrac{b}{c}\\ \Rightarrow10ac+bc=10b^2+bc\\ \Rightarrow ac=b^2\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\Rightarrow\dfrac{a^2}{b^2}=\dfrac{b^2}{c^2}=\dfrac{a^2+b^2}{b^2+c^2}\\ \text{Đặt }\dfrac{a}{b}=\dfrac{b}{c}=k\Rightarrow a=bk;b=ck\\ \Rightarrow a=bk=ck^2\Rightarrow\dfrac{a}{c}=k^2=\dfrac{a^2}{b^2}\\ \Rightarrow\dfrac{a}{c}=\dfrac{a^2+b^2}{b^2+c^2}\)
Vế sau ko cm được vì ko có ẩn d trong đề bài
