a. TH1:

\(\left\{{}\begin{matrix}x^2+3x-4< 0\\3-2x>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>-4\end{matrix}\right.\\x>\dfrac{3}{2}\end{matrix}\right.\)

TH2:

\(\left\{{}\begin{matrix}x^2+3x-4>0\\3-2x< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x< -4\end{matrix}\right.\\x< \dfrac{3}{2}\end{matrix}\right.\)

Vậy nghiệm của BPT:

\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>-4\end{matrix}\right.\\x>\dfrac{3}{2}\end{matrix}\right.\)      \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x< -4\end{matrix}\right.\\x< \dfrac{3}{2}\end{matrix}\right.\)

\(a,f'\left(x\right)=3x^2-6x\\ f'\left(x\right)\le0\Leftrightarrow3x^2-6x\le0\\ \Leftrightarrow3x\left(x-2\right)\le0\Leftrightarrow0\le x\le2\)

Lời giải:

a. $f'(x)\leq 0$

$\Leftrightarrow 3x^2-6x\leq 0$

$\Leftrightarrow x(x-2)\leq 0$

$\Leftrightarrow 0\leq x\leq 2$

b.

$f'(x)=x^2-3x+2=0$

$\Leftrightarrow 3x^2-6x=x^2-3x+2=0$

$\Leftrightarrow 3x(x-2)=(x-1)(x-2)=0$

$\Leftrightarrow x-2=0$

$\Leftrightarrow x=2$

c.

$g(x)=f(1-2x)+x^2-x+2022$

$g'(x)=(1-2x)'f(1-2x)'_{1-2x}+2x-1$

$=-2[3(1-2x)^2-6(1-2x)]+2x-1$
$=-24x^2+2x+5$

$g'(x)\geq 0$

$\Leftrightarrow -24x^2+2x+5\geq 0$

$\Leftrightarrow (5-12x)(2x-1)\geq 0$

$\Leftrightarrow \frac{-5}{12}\leq x\leq \frac{1}{2}$

tách nhỏ câu hỏi ra nhé dài quá

a) \(\left(x+1\right)\left(x-1\right)\left(3x-6\right)>0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow\left[{}\begin{matrix}-1< x< 1\\x>2\end{matrix}\right.\)

b) \(\dfrac{x+3}{x-2}\le0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow-3\le x< 2\)

d) \(\dfrac{2x-5}{3x+2}< \dfrac{3x+2}{2x-5}\)

\(\Leftrightarrow\dfrac{2x-5}{3x+2}-\dfrac{3x+2}{2x-5}< 0\)

\(\Leftrightarrow\dfrac{\left(2x-5\right)^2-\left(3x+2\right)^2}{\left(3x+2\right)\left(2x-5\right)}< 0\)

\(\Leftrightarrow\dfrac{\left(2x-5+3x+2\right)\left(2x-5-3x-2\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)

\(\Leftrightarrow\dfrac{-\left(5x-3\right)\left(x+7\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow\left[{}\begin{matrix}-7< x< -\dfrac{2}{3}\\\dfrac{5}{3}< x< \dfrac{5}{2}\end{matrix}\right.\)

a: =>2x^2+8x-3x-12<2x^2+2

=>5x<14

=>x<14/5

b: =>\(\dfrac{9x-3-\left(5x+1\right)\left(x-2\right)}{3\left(x-2\right)}-4>0\)

=>\(\dfrac{9x-3-5x^2+10x-x+2-12\left(x-2\right)}{3\left(x-2\right)}>0\)

=>\(\dfrac{-5x^2+18x-1-12x+24}{3\left(x-2\right)}>0\)

=>\(\dfrac{-5x^2+6x+23}{x-2}>0\)

TH1: x-2>0 và -5x^2+6x+23>0

=>x>2 và \(\dfrac{3-2\sqrt{31}}{5}< x< \dfrac{3+2\sqrt{31}}{5}\)

=>\(2< x< \dfrac{3+2\sqrt{31}}{5}\)

TH2: x-2<0 và -5x^2+6x+23<0

=>x<2 và \(\left[{}\begin{matrix}x< \dfrac{3-2\sqrt{31}}{5}\\x>\dfrac{3+2\sqrt{31}}{5}\end{matrix}\right.\)

=>\(x< \dfrac{3-2\sqrt{31}}{5}\)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne1\\x\ne2\\x\ne7\end{matrix}\right.\)

\(\dfrac{2\left(x-4\right)}{\left(x-1\right)\left(x-7\right)}\ge\dfrac{1}{x-2}\\ \Leftrightarrow\dfrac{2x-8}{x^2-8x+7}\ge\dfrac{1}{x-2}\\ \Leftrightarrow\left(2x-8\right)\left(x-2\right)\ge x^2-8x+7\)

\(\Leftrightarrow2x^2-12x+16\ge x^2-8x+7\\ \Leftrightarrow x^2-4x+9\ge0\left(luôn.đúng\right)\)