Question

giải PT

4y²+1=4y

y²-2y=80

 

Answer 1

\(4y^2+1=4y\\ \Leftrightarrow4y^2+1-4y=0\\ \Leftrightarrow4y^2-4y+1=0\\ \Leftrightarrow2y\left(2y-1\right)-\left(2y-1\right)=0\\ \Leftrightarrow\left(2y-1\right)^2=0\\ \Leftrightarrow2y-1=0\\ \Leftrightarrow y=\dfrac{1}{2}\)

Answer 2

\(y^2-2y=80\\ \Leftrightarrow y^2-2y-80=0\\ \Leftrightarrow y^2+8y-10y-80=0\\ \Leftrightarrow y\left(y+8\right)-10\left(y+8\right)=0\\ \Leftrightarrow\left(y+8\right)\left(y-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}y=-8\\x=10\end{matrix}\right.\)