\(\left(2x-1\right)^2=49\)
<=>\(\left(2x-1\right)^2=7^2\)
<=>\(2x-1=7\)
<=>\(2x=8\)
<=>\(x=4\)
\(\left(5x-3\right)^2-\left(4x-7\right)^2=0\)
<=>\(\orbr{\begin{cases}5x-3=0\\4x-7=0\end{cases}}\)
<=>\(\orbr{\begin{cases}5x=3\\4x=7\end{cases}}\)
<=>\(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{7}{4}\end{cases}}\)
\(\left(2x-1\right)^2=49\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=7\\2x-1=-7\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=8\\2x=-6\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=-3\end{cases}}}\)
Vậy x=4; x=-3
x= 4 x= 3
HOK TOT!!!
