a) \(4\sqrt{x}+\frac{2}{\sqrt{x}}< 2x+\frac{1}{2x}+2\)

hay \(2\sqrt{x}+\frac{1}{\sqrt{x}}< x+\frac{1}{4x}+1\)

\(\Leftrightarrow0< x+\frac{1}{4x}+1-2\sqrt{x}-\frac{1}{\sqrt{x}}\)

\(\Leftrightarrow0< \left(\sqrt{x}\right)^2-2\sqrt{x}-2\sqrt{x}\cdot1+1+\frac{1}{\left(2\sqrt{x}\right)^2}-2\cdot\frac{1}{2\sqrt{x}}\)

\(\Leftrightarrow1< \left(\sqrt{x}-1\right)^2+\left(\frac{1}{2\sqrt{x}}-1\right)^2\)

\(\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}>1\\2\sqrt{x}>1\end{cases}\Rightarrow\hept{\begin{cases}x>1\\x>\frac{1}{4}\end{cases}\Rightarrow}x>1}\)

b) \(\frac{1}{1-x^2}>\frac{3}{\sqrt{1-x^2}}-1\left(1\right)\left(ĐK:-1< x< 1\right)\)

Ta có (1) <=> \(\frac{1}{1-x^2}-1-\frac{3x}{\sqrt{1-x^2}}+2>0\)\(\Leftrightarrow\frac{x^2}{1-x^2}-\frac{3x}{\sqrt{1-x^2}}+2>0\)

Đặt \(t=\frac{x}{\sqrt{1-x^2}}\)ta được

\(t^2-3t+2>0\Leftrightarrow\orbr{\begin{cases}\frac{x}{\sqrt{1-x^2}}< 1\\\frac{x}{\sqrt{1-x^2}}>2\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{1-x^2}>x\left(a\right)\\2\sqrt{1-x^2}< x\left(b\right)\end{cases}}}\)

(a) <=> \(\hept{\begin{cases}x< 0\\1-x^2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge0\\1-x^2>x^2\end{cases}}}\)

\(\Leftrightarrow-1< x< 0\)hoặc \(\hept{\begin{cases}x\ge0\\x^2< \frac{1}{2}\end{cases}}\)

\(\Leftrightarrow-1< x< 0\)hoặc \(0\le x\le\frac{\sqrt{2}}{2}\Leftrightarrow-1< x< \frac{\sqrt{2}}{2}\)

(b) \(\Leftrightarrow\hept{\begin{cases}1-x^2>0\\x>0\\4\left(1-x^2\right)< x^2\end{cases}\Leftrightarrow\hept{\begin{cases}0< x< 1\\x^2>\frac{4}{5}\end{cases}\Leftrightarrow}\frac{2}{\sqrt{5}}< x< 1}\)

ok đợi nấu ăn xong r làm cho

2/ \(\left[{}\begin{matrix}x< -12\\x>12\end{matrix}\right.\)

- Với \(x< -12\Rightarrow x+\frac{12x}{\sqrt{x^2-144}}=x\left(1+\frac{12}{\sqrt{x^2-144}}\right)< 0< 35\)

\(\Rightarrow\) BPT luôn đúng

- Với \(x>12\), hai vế không âm, bình phương hai vế ta được:

\(x^2+\frac{144x^2}{x^2-144}+24\frac{x^2}{\sqrt{x^2-144}}-1225\le0\)

\(\Leftrightarrow\frac{x^4}{x^2-144}+24\frac{x^2}{\sqrt{x^2-144}}-1225\le0\)

\(\Leftrightarrow\left(\frac{x^2}{\sqrt{x^2-144}}+49\right)\left(\frac{x^2}{\sqrt{x^2-144}}-25\right)\le0\)

\(\Leftrightarrow\frac{x^2}{\sqrt{x^2-144}}-25\le0\)

\(\Leftrightarrow x^2\le25\sqrt{x^2-144}\)

\(\Leftrightarrow x^4-625x^2+90000\le0\)

\(\Leftrightarrow\left(x^2-400\right)\left(x^2-225\right)\le0\)

\(\Leftrightarrow225\le x^2\le400\)

\(\Leftrightarrow15\le x\le20\)

Vậy nghiệm của BPT là \(\left[{}\begin{matrix}x< -12\\15\le x\le20\end{matrix}\right.\)