Bài 2:
a) Hàm số đi qua điểm `A(1/2;1/4)` thay `x=1/2;y=1/4` ta có:
\(\dfrac{1}{4}=\left(3m+1\right)\cdot\left(\dfrac{1}{2}\right)^2\)
\(\Leftrightarrow\left(3m+1\right)\cdot\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow3m+1=1\)
\(\Leftrightarrow3m=0\)
\(\Leftrightarrow m=0\) (tm)
b) \(\left\{{}\begin{matrix}3x-4y=2\\-4x+3y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12x-16y=8\\-12x+9y=-15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-7y=-7\\-4x+3y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\-4x+3=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\-4x=-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)
\(\Rightarrow B\left(2;1\right)\)
Thay x;y vào hàm số ta có:
\(1=\left(3m+1\right)\cdot2^2\)
\(\Leftrightarrow\left(3m+1\right)\cdot4=1\)
\(\Leftrightarrow3m+1=\dfrac{1}{4}\)
\(\Leftrightarrow3m=-\dfrac{3}{4}\)
\(\Leftrightarrow m=-\dfrac{3}{4}:3\)
\(\Leftrightarrow m=-\dfrac{1}{4}\)(tm)
Ai giúp em với ạ em cần rất gấp bài 4 thôi ạ
