Question

2 Tìm x

a) 49x²-1=0

b) (2x-1)²-(4x+1) (x-3)=-3

Answer 1

a,49x²-1=0

\(\Leftrightarrow\left(7x\right)^2-1^2=0\)\(\Leftrightarrow\left(7x-1\right)\left(7x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-1=0\\7x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=1\\7x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=\dfrac{-1}{7}\end{matrix}\right.\)

b,(2x-1)²-(4x+1)(x-3)= -3

\(\Leftrightarrow\left[\left(2x\right)^2-2x+1\right]-\left(4x^2-12x+x-3\right)=0\)

\(\Leftrightarrow4x^2-2x+1-4x^2+12x-x+3=0\)

\(\Leftrightarrow9x+4=0\Leftrightarrow9x=-4\Leftrightarrow x=\dfrac{-4}{9}\)