Question

[Trần Khang]

4) \(\dfrac{x-\sqrt{x}}{1-\sqrt{2\left(x^2-x+1\right)}}\ge1\)

5)\(x^2+x+1>3\sqrt{x}\left(x+1\right)\)

6)\(\dfrac{1}{1-x^2}>\dfrac{3x}{\sqrt{1-x^2}}-1\)

nữa ạ

Answer 1

[Hồng Phúc]

4.

ĐK: \(x\ge0\)

Ta có \(1-\sqrt{2\left(x^2-x+1\right)}\le1-\sqrt{2}< 0\), khi đó:

\(\dfrac{x-\sqrt{x}}{1-\sqrt{2\left(x^2-x+1\right)}}\ge1\)

\(\Leftrightarrow x-\sqrt{x}\le1-\sqrt{2\left(x^2-x+1\right)}\)

\(\Leftrightarrow\sqrt{x}-\dfrac{1}{\sqrt{x}}-1+\sqrt{2\left(x+\dfrac{1}{x}-1\right)}\le0\)

\(\Leftrightarrow\sqrt{x}-\dfrac{1}{\sqrt{x}}-1+\sqrt{2\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)^2+2}\le0\)

\(\Leftrightarrow t-1+\sqrt{2t^2+2}\le0\left(t=\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)

\(\Leftrightarrow\sqrt{2t^2+2}\le1-t\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-t>0\\2t^2+2\le t^2-2t+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t< 1\\\left(t+1\right)^2\le0\end{matrix}\right.\)

\(\Leftrightarrow t=-1\)

\(\Leftrightarrow\sqrt{x}-\dfrac{1}{\sqrt{x}}+1=0\)

\(\Leftrightarrow x+\sqrt{x}-1=0\)

\(\Leftrightarrow x=\dfrac{3-\sqrt{5}}{2}\)