Question

1) \(\sqrt{2x^2+4x-1}>x+1\)

2)\(\sqrt{4x^2+101x+64}>2\left(x+10\right)\)

3)\(\sqrt{x^2-5x-14}>x-3\)

Answer 1

\(\sqrt{2x^2+4x-1}=\sqrt{2\left(x+\dfrac{2-\sqrt{6}}{2}\right)\left(x+\dfrac{2+\sqrt{6}}{2}\right)}\)

\(\Rightarrow\) BXD :

\(x\)	\(-\infty\)		\(\dfrac{-2-\sqrt{6}}{2}\)		\(\dfrac{-2+\sqrt{6}}{2}\)		\(+\infty\)
\(x+\dfrac{2-\sqrt{6}}{2}\)	\(-\)	\(-\)	\(-\)	\(-\)	\(0\)	\(+\)	\(+\)
\(x+\dfrac{2+\sqrt{6}}{2}\)	\(-\)	\(-\)	\(0\)	\(+\)	\(+\)	\(+\)	\(+\)
\(\sqrt{2x^2+4x-1}\)	\(+\)	\(+\)	\(0\)	oxđ	\(0\)	\(+\)	\(+\)

\(\Rightarrow\) \(\sqrt{2x^2+4x-1}\) xác định \(\Leftrightarrow x\in\left(-\infty;\dfrac{-2-\sqrt{6}}{2}\right)\cup\left(\dfrac{-2+\sqrt{6}}{2};+\infty\right)\)

ta có : \(\sqrt{2x^2+4x-1}>x+1\)

\(\Leftrightarrow\sqrt{2x^2+4x-1}-x-1>0\)

\(\Rightarrow\) BXD :

\(x\)	\(-\infty\)		\(\dfrac{-2-\sqrt{6}}{2}\)		\(-1\)		\(\dfrac{-2+\sqrt{6}}{2}\)		\(+\infty\)
\(\sqrt{2x^2+4x-1}\)	\(+\)	\(+\)	\(0\)	oxđ	oxđ	oxđ	\(0\)		\(+\)
\(-x-1\)	\(+\)	\(+\)	\(\dfrac{\sqrt{6}}{2}\)	bỏ	\(0\)	bỏ	\(\dfrac{-\sqrt{6}}{2}\)		\(-\)
\(f\left(x\right)\)	\(+\)	\(+\)	\(0\)	bỏ		bỏ	\(0\)		không rỏ dấu

bn nào giỏi lm tiếp đi nha