Cho (d) y=(4m-3)x+3m+4. Tìm m để (d) cắt trục Ox và Oy tại A;B sao cho tam giác OAB cân

\(x=0→y=3m+4→OB=|3m+4|\\y=0→x=\dfrac{-3m-4}{4m-3}→OA=\bigg|\dfrac{-3m-4}{4m-3}\bigg|\)

Để \(ΔOAB\) vuông cân

\(→OA=OB\) hay \(\bigg|\dfrac{-3m-4}{4m-3}\bigg|=|3m+4|(m\ne \dfrac{3}{4})\)

\(↔\left[\begin{array}{1}\dfrac{-3m-4}{4m-3}=3m+4\\\dfrac{-3m-4}{4m-3}=-3m-4\end{array}\right.\\↔\left[\begin{array}{1}-3m-4=(3m+4)(4m-3)\\-3m-4=(-3m-4)(4m-3)\end{array}\right.\\↔\left[\begin{array}{1}-3m-4=12m^2+7m-12\\-3m-4=-12m^2-7m+12\end{array}\right.\\↔\left[\begin{array}{1}-12m^2-10m+8=0\\12m^2+4m-16=0\end{array}\right.\\↔\left[\begin{array}{1}6m^2+5m-4=0\\3m^2+m-4=0\end{array}\right.\\↔\left[\begin{array}{1}6m^2+8m-3m-4=0\\3m^2+4m-3m-4=0\end{array}\right.\\↔\left[\begin{array}{1}2m(3m+4)-(3m+4)=0\\m(3m+4)-(3m+4)=0\end{array}\right.\\↔\left[\begin{array}{1}(2m-1)(3m+4)=0\\(m-1)(3m+4)=0\end{array}\right.\\↔\left[\begin{array}{1}2m-1=0\\3m+4=0\\m-1=0\end{array}\right.\\↔\left[\begin{array}{1}m=\dfrac{1}{2}(TM)\\m=-\dfrac{4}{3}(TM)\\m=1(TM)\end{array}\right.\)

Vậy \(m∈\{\dfrac{1}{2};-\dfrac{4}{3};1\}\)