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

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.

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

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

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

$g'(x)\geq 0$

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

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

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

\(\left\{{}\begin{matrix}x^2-3x+2\le0\\mx+1-m\le0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}1\le x\le2\\x\le\dfrac{-1+m}{m}\end{matrix}\right.\)

để hpt trên có nghiệm thì \(\dfrac{-1+m}{m}\le2\) ĐK m ≠ 0

\(< =>m\ge-1\)

Vậy .....

\(x^2-3x+2\le0\Leftrightarrow1\le x\le2\) \(\Rightarrow D_1=\left[1;2\right]\)

Xét \(mx\le m-1\)

- Với \(m=0\) BPT vô nghiệm

- Với \(m>0\Leftrightarrow x\le\dfrac{m-1}{m}\) \(\Rightarrow D_2=(-\infty;\dfrac{m-1}{m}]\)

Hệ có nghiệm khi \(D_1\cap D_2\ne\varnothing\)

\(\Leftrightarrow\dfrac{m-1}{m}\ge1\) \(\Rightarrow\) không tồn tại m thỏa mãn

- Với \(m< 0\Leftrightarrow x\ge\dfrac{m-1}{m}\Rightarrow D_2=[\dfrac{m-1}{m};+\infty)\)

\(D_1\cap D_2\ne\varnothing\Leftrightarrow\dfrac{m-1}{m}\le2\)

\(\Leftrightarrow m-1\ge2m\Rightarrow m\le-1\)

Vậy \(m\le-1\)

a) ĐKXĐ : x khác 2/5

\(\frac{2x+3}{2-5x}\le0\)

\(\Leftrightarrow2x+3\le2-5x\)

\(\Leftrightarrow7x\le-1\)

\(\Leftrightarrow x\le\frac{-1}{7}\left(\text{thỏa mãn}\right)\)

b) \(\left|5x+3\right|=\left|x+2\right|\)

\(\Rightarrow\orbr{\begin{cases}5x+3=x+2\\5x+3=-x-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}4x=-1\\6x=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=\frac{-5}{6}\end{cases}}\)

P.s: cái này chưa học có j sai góp ý hộ nha ^^

ĐKXĐ: \(x\ne\frac{2}{5}\)

\(\frac{2x+3}{x-5x}\le0\)

Xét 2 trường hợp

TH1: \(\hept{\begin{cases}2x+3\ge0\\2-5x< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge-\frac{3}{2}\\x>\frac{2}{5}\end{cases}}}\Leftrightarrow x>\frac{2}{5}\)

TH2: \(\hept{\begin{cases}2x+3\le0\\2-5x>0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le-\frac{3}{2}\\x< \frac{2}{5}\end{cases}}}\Leftrightarrow x\le-\frac{3}{2}\)

Vậy \(\orbr{\begin{cases}x>\frac{2}{5}\\x\le-\frac{3}{2}\end{cases}}\)

P/S: chưa học => trình bày thiếu sót ( sai ) => sửa hộ~

vfhgjhgjhhgjfdgfhgfrdtfytrftrrtfytrfyt

Bài 1 \(\left\{{}\begin{matrix}x^2-3x-4\le0\\\left(m-1\right)x\ge2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1\le x\le4\\\left(m-1\right)x\ge2\end{matrix}\right.\)

Nếu m = 1, hệ vô nghiệm

Nếu m ≠ 1, hệ tương đương

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}-1\le m< 1\\x\le\dfrac{2}{m-1}\end{matrix}\right.\\\left\{{}\begin{matrix}1< m\le4\\x\ge\dfrac{2}{m-1}\end{matrix}\right.\end{matrix}\right.\)

Hệ có nghiệm khi một trong hai hệ trong hệ ngoặc vuông có nghiệm ⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}-1\le m< 1\\\dfrac{2}{m-1}\ge-1\end{matrix}\right.\\\left\{{}\begin{matrix}1< m\le4\\\dfrac{2}{m-1}\le4\end{matrix}\right.\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}-1\le m< 1\\-2\le1-m\end{matrix}\right.\\\left\{{}\begin{matrix}1< m\le4\\2\le4m-4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-1\le m< 1\\\dfrac{3}{2}\le m\le4\end{matrix}\right.\)

a/

\(\Leftrightarrow\frac{\left(x^2-1\right)\left(x^2+1\right)}{x^2+3x}+x^2-1\ge0\)

\(\Leftrightarrow\left(x^2-1\right)\left(\frac{x^2+1}{x^2+3x}+1\right)\ge0\)

\(\Leftrightarrow\left(x^2-1\right)\left(\frac{2x^2+3x+1}{x^2+3x}\right)\ge0\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(2x+1\right)}{x\left(x+3\right)}\ge0\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(2x+1\right)\left(x+1\right)^2}{x\left(x+3\right)}\ge0\)

\(\Rightarrow\left[{}\begin{matrix}x< -3\\x=-1\\-\frac{1}{2}\le x< 0\\x\ge1\end{matrix}\right.\)

b/

\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)\left(\frac{-2-2x}{x}\right)\le0\)

\(\Leftrightarrow\frac{-2.\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+1\right)}{x}\le0\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(x-1\right)\left(x-2\right)\left(x+1\right)^2}{x}\ge0\)

\(\Rightarrow\left[{}\begin{matrix}x\le-2\\x=-1\\0< x\le1\\x\ge2\end{matrix}\right.\)

c/

\(\Leftrightarrow\left(\frac{4\left(x-1\right)-2x}{x\left(x-1\right)}\right)\left(\frac{x^2+1-2x}{x}\right)\le0\)

\(\Leftrightarrow\frac{\left(2x-4\right)\left(x-1\right)^2}{x^2\left(x-1\right)}\le0\)

\(\Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)^2}{x^2\left(x-1\right)}\le0\)

\(\Rightarrow1< x\le2\)

d/

ĐKXĐ: \(\left\{{}\begin{matrix}x^3-4x\ge0\\\frac{1+x}{x}-2\ge0\\x\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x-2\right)\left(x+2\right)\ge0\\\frac{1-x}{x}\ge0\\x\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}-2\le x\le0\\x\ge2\end{matrix}\right.\\0< x\le1\\x\ne0\end{matrix}\right.\)

\(\Rightarrow\) Không tồn tại x thỏa mãn ĐKXĐ

Vậy BPT đã cho vô nghiệm

7,3, -6

ĐKXĐ: \(x\ne7;x\ne2\)

BPT \(\Leftrightarrow f\left(x\right)=\dfrac{\left(6-2x\right)^3\left(x+6\right)}{\left(x-7\right)^3}\le0\)

Lập bảng xét dấu ta có:

Từ đây ta thấy \(-6\le x\le3\) hoặc \(x>7\) thỏa mãn bất phương trình ban đầu.

Vậy...