Question

help m.n ơi 

Answer 1

\(49x^2-81=0\)
\(49x^2=81\)
\(x^2=\dfrac{81}{49}\)
\(=>x=\sqrt{\dfrac{81}{49}}=\dfrac{9}{7}\)
\(\dfrac{x-3}{4}=\dfrac{x}{3}\)
\(3x-9=4x\)
\(x=-9\)
 

Answer 2

a)\(49x^2-81=0\\ \Leftrightarrow\left(7x-9\right)\left(7x+9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7x-9=0\\7x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{-9}{7}\end{matrix}\right.\)

b)\(\dfrac{x-3}{4}=\dfrac{x}{3}\Leftrightarrow3\left(x-3\right)=4x\Leftrightarrow3x-9=4x\Leftrightarrow x=-9\)