Question

giải phương trình: \(\sqrt{4x^2+4x+1}=6\)

Answer 1

\(\sqrt{4x^2+4x+1}=6\Leftrightarrow\sqrt{\left(2x\right)^2+2.2x.1+1}=6\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\Leftrightarrow\left|2x+1\right|=6\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy S={\(-\dfrac{7}{2};\dfrac{5}{2}\)}

Answer 2

\(\Leftrightarrow4x^2+4x+1=36\)

\(\Leftrightarrow4x^2+4x-35=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)