a: \(\Leftrightarrow2x^2-2-3>-5x+\left(2x+1\right)\left(x-3\right)\)

\(\Leftrightarrow2x^2-5>-5x+2x^2-6x+x-3\)

\(\Leftrightarrow2x^2-5>2x^2-10x-3\)

=>-5>-10x-3

=>5<10x+3

=>10x+3>5

=>10x>2

hay x>1/5

b: \(\Leftrightarrow x^2-6x+9+8-4x>x+7\)

\(\Leftrightarrow x^2-10x+17-x-7>0\)

\(\Leftrightarrow x^2-11x+10>0\)

=>x>10 hoặc x<1

a: 

=>-5>-10x-3

=>5<10x+3

=>10x+3>5

=>10x>2

hay x>1/5

b: 

=>x>10 hoặc x<1

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-\dfrac{9}{2}\\x\ne0\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{\left(3+\sqrt{9+2x}\right)^2.2x^2}{\left(3-\sqrt{9+2x}\right)^2\left(3+\sqrt{9+2x}\right)^2}< x+21\)

\(\Leftrightarrow\dfrac{\left(3+\sqrt{9+2x}\right)^2.2x^2}{4x^2}< x+21\)

\(\Leftrightarrow\left(3+\sqrt{9+2x}\right)^2< 2x+42\)

\(\Leftrightarrow x+9+3\sqrt{9+2x}< x+21\)

\(\Leftrightarrow\sqrt{9+2x}< 4\)

\(\Leftrightarrow9+2x< 16\Rightarrow x< \dfrac{7}{2}\)

Vậy \(\left\{{}\begin{matrix}-\dfrac{9}{2}\le x< \dfrac{7}{2}\\x\ne0\end{matrix}\right.\)

ĐKXĐ: \(x\ge-\frac{3}{2}\)

Do \(1+\sqrt{3+2x}>0\) nên BPT tương đương:

\(4\left(x+1\right)^2\left(1+\sqrt{3+2x}\right)^2< \left(2x+1\right)\left(1-\sqrt{3+2x}\right)^2\left(1+\sqrt{3+2x}\right)^2\)

\(\Leftrightarrow4\left(x+1\right)^2\left(1+\sqrt{3+2x}\right)^2< \left(2x+1\right).4\left(x+1\right)^2\)

- Với \(x=-1\) ko phải là nghiệm

- Với \(x\ne-1\)

\(\Leftrightarrow\left(1+\sqrt{3+2x}\right)^2< 2x+1\)

\(\Leftrightarrow4+2x+2\sqrt{3+2x}< 2x+1\)

\(\Leftrightarrow2\sqrt{3+2x}< -3\)

BPT vô nghiệm

\(f'\left(x\right)=\dfrac{1-x}{\sqrt{2x-x^2}}\)

\(f'\left(x\right)\ge1\Leftrightarrow\dfrac{1-x}{\sqrt{2x-x^2}}\ge1\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-x^2>0\\1-x>0\\\left(1-x\right)^2\ge2x-x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}0< x< 2\\x< 1\\2x^2-4x+1\ge0\end{matrix}\right.\) \(\Rightarrow0< x\le\dfrac{2-\sqrt{2}}{2}\)

f^'(x)=\(\dfrac{2-2x}{2\sqrt{2x-x^2}}\) = \(\dfrac{1-x}{\sqrt{2x-x^2}}\)

để f^'(x) \(\ge\) 1 \(\Leftrightarrow\) \(\dfrac{1-x}{\sqrt{2x-x^2}}\) \(\ge\) 1 \(\Leftrightarrow\) 1-x \(\ge\) \(\sqrt{2x-x^2}\) 

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x-x^2>0\\1-2x+x^2\ge2x-x^2\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}0< x< 2\\\left\{{}\begin{matrix}x< \dfrac{2-\sqrt{2}}{2}\\x>\dfrac{2+\sqrt{2}}{2}\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\) 0<x\(\le\) \(\dfrac{2-\sqrt{2}}{2}\)

\(\left(2x+1\right)^2+\left(1-x\right)3x\le\left(x+2\right)^2\)

\(\Leftrightarrow4x^2+4x+1+3x-3x^2\le x^2+4x+4\)

\(\Leftrightarrow4x^2+4x+3x-3x^2-x^2-4x\le4-1\)

\(\Leftrightarrow3x\le3\Leftrightarrow x\le1\) vậy \(x\le1\)