Question

giải phương trình vô tỉ

\(\sqrt{4x^2+4x+1}\) = 6

Answer 1

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

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

Answer 2

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow|2x+1|=6\)
\(\Leftrightarrow2x+1=6\)
\(\Leftrightarrow2x=5 \)
\(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy nghiệm của phương trình S=\(\left\{\dfrac{5}{2}\right\}\)