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

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.

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

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

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

$g'(x)\geq 0$

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

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

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

\(\dfrac{2x-1}{x+1}-2< 0.\left(x\ne-1\right).\\ \Leftrightarrow\dfrac{2x-1-2x-2}{x+1}< 0.\Leftrightarrow\dfrac{-3}{x+1}< 0.\)

Mà \(-3< 0.\)

\(\Rightarrow x+1>0.\Leftrightarrow x>-1\left(TMĐK\right).\)

\(\dfrac{x^2-2x+5}{x-2}-x+1\ge0.\left(x\ne2\right).\\ \Leftrightarrow\dfrac{x^2-2x+5-x^2+2x+x-2}{x-2}\ge0.\\ \Leftrightarrow\dfrac{x+3}{x-2}\ge0.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0.\\x-2\ge0.\end{matrix}\right.\\\left\{{}\begin{matrix}x+3\le0.\\x-2\le0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3.\\x\ge2.\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3.\\x\le2.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge2.\\x\le-3.\end{matrix}\right.\)

Kết hợp ĐKXĐ.

\(\Rightarrow\left[{}\begin{matrix}x>2.\\x\le-3.\end{matrix}\right.\)

\(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}\le0.\left(x\ne1;x\ne\dfrac{-3}{2}\right).\)

Đặt \(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}=f\left(x\right).\)

Ta có bảng sau:

Vậy \(f\left(x\right)\ge0.\Leftrightarrow x\in\left(\dfrac{-3}{2};\dfrac{-1}{2}\right)\cup\)(1;2].

2) −(x2−2x+5)x−2−x+1≥0−(x2−2x+5)x−2−x+1≥0                                                        ⇔$-{}^{}$

1: \(\Leftrightarrow\left[{}\begin{matrix}2x-3>5\\2x-3< -5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>5\\x< -1\end{matrix}\right.\)

2: \(\Leftrightarrow-4< =2x-1< =4\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1>=-4\\2x-1< =4\end{matrix}\right.\Leftrightarrow\dfrac{-3}{2}< =x< =\dfrac{5}{2}\)

a, ĐK: \(x-2>0\Rightarrow x>2\)

\(log_2\left(x-2\right)< 2\\ \Leftrightarrow x-2< 4\\ \Leftrightarrow x< 6\)

Kết hợp với ĐKXĐ, ta được: \(2< x< 6\)

b, ĐK: \(2x-1>0\Leftrightarrow x>\dfrac{1}{2}\)

\(log\left(x+1\right)\ge log\left(2x-1\right)\\ \Leftrightarrow x+1\ge2x-1\\ \Leftrightarrow x\le2\)

Kết hợp với ĐKXĐ, ta được: \(\dfrac{1}{2}< x\le2\)

a) \({\log _{\frac{1}{7}}}\left( {x + 1} \right) > {\log _7}\left( {2 - x} \right)\)               (ĐK: \(x + 1 > 0;2 - x > 0 \Leftrightarrow  - 1 < x < 2\))

\(\begin{array}{l} \Leftrightarrow {\log _{{7^{ - 1}}}}\left( {x + 1} \right) > {\log _7}\left( {2 - x} \right)\\ \Leftrightarrow  - {\log _7}\left( {x + 1} \right) > {\log _7}\left( {2 - x} \right)\\ \Leftrightarrow {\log _7}{\left( {x + 1} \right)^{ - 1}} > {\log _7}\left( {2 - x} \right)\\ \Leftrightarrow {\left( {x + 1} \right)^{ - 1}} > 2 - x\\ \Leftrightarrow \frac{1}{{x + 1}} - 2 + x > 0\\ \Leftrightarrow \frac{{1 + \left( {x - 2} \right)\left( {x + 1} \right)}}{{x + 1}} > 0\\ \Leftrightarrow \frac{{1 + {x^2} - x - 2}}{{x + 1}} > 0 \Leftrightarrow \frac{{{x^2} - x - 1}}{{x + 1}} > 0\end{array}\)

Mà – 1 < x < 2 nên x + 1 > 0

\( \Leftrightarrow {x^2} - x - 1 > 0 \Leftrightarrow \left[ \begin{array}{l}x < \frac{{1 - \sqrt 5 }}{2}\\x > \frac{{1 + \sqrt 5 }}{2}\end{array} \right.\)

KHĐK ta có \(\left[ \begin{array}{l} - 1 < x < \frac{{1 - \sqrt 5 }}{2}\\\frac{{1 + \sqrt 5 }}{2} < x < 2\end{array} \right.\)

b) \(2\log \left( {2x + 1} \right) > 3\)              (ĐK: \(2x + 1 > 0 \Leftrightarrow x > \frac{{ - 1}}{2}\))

\(\begin{array}{l} \Leftrightarrow \log \left( {2x + 1} \right) > \frac{3}{2}\\ \Leftrightarrow 2x + 1 > {10^{\frac{3}{2}}} = 10\sqrt {10} \\ \Leftrightarrow x > \frac{{10\sqrt {10}  - 1}}{2}\end{array}\)

KHĐK ta có \(x > \frac{{10\sqrt {10}  - 1}}{2}\)

4: =>2x-3>5 hoặc 2x-3<-5

=>x>4 hoặc x<-1

5: =>-4<=2x-1<=4

=>-3/2<=x<=5/2

a: \(log\left(x-2\right)< 3\)

=>\(\left\{{}\begin{matrix}x-2>0\\log\left(x-2\right)< log9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-2>0\\x-2< 9\end{matrix}\right.\Leftrightarrow2< x< 11\)

b: \(log_2\left(2x-1\right)>3\)

=>\(\left\{{}\begin{matrix}2x-1>0\\log_2\left(2x-1\right)>log_29\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-1>0\\2x-1>9\end{matrix}\right.\Leftrightarrow2x-1>9\)

=>2x>10

=>x>5

c: \(log_3\left(-x-1\right)< =2\)

=>\(\left\{{}\begin{matrix}-x-1>0\\log_3\left(-x-1\right)< =log_39\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-x-1>0\\-x-1< =9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x>1\\-x< =10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< -1\\x>=-10\end{matrix}\right.\Leftrightarrow-10< =x< -1\)

d: \(log_2\left(2x-3\right)>=2\)

=>\(\left\{{}\begin{matrix}2x-3>0\\log_2\left(2x-3\right)>=log_24\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-3>0\\2x-3>=4\end{matrix}\right.\)

=>2x-3>=4

=>2x>=7

=>\(x>=\dfrac{7}{2}\)

e: \(log_3\left(2x-7\right)>2\)

=>\(\left\{{}\begin{matrix}2x-7>0\\log_3\left(2x-7\right)>log_39\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>\dfrac{7}{2}\\2x-7>9\end{matrix}\right.\)

=>2x-7>9

=>2x>16

=>x>8

a.

\(log\left(x-2\right)< 3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2>0\\x-2< 10^3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< 1002\end{matrix}\right.\) \(\Rightarrow2< x< 1002\)

b.

\(log_2\left(2x-1\right)>3\Leftrightarrow\left\{{}\begin{matrix}2x-1>0\\2x-1>2^3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{2}\\x>\dfrac{9}{2}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{9}{2}\)

c.

\(log_3\left(-x-1\right)\le2\Rightarrow\left\{{}\begin{matrix}-x-1>0\\-x-1\le3^2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x< -1\\x\ge-10\end{matrix}\right.\) \(\Rightarrow-10\le x< -1\)

d.

\(log_2\left(2x-3\right)\ge2\Leftrightarrow\left\{{}\begin{matrix}2x-3>0\\2x-3\ge2^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x>\dfrac{7}{2}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{7}{2}\)

e,

\(log_3\left(2x-7\right)>2\Leftrightarrow\left\{{}\begin{matrix}2x-7>0\\2x-7>3^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{7}{2}\\x>8\end{matrix}\right.\) \(\Rightarrow x>8\)

Lời giải:
a. ĐK: $x>2$
$\log(x-2)<3$

$\Leftrightarrow x-2< 10^3$

$\Leftrightarrow x< 1002$

Vậy $2< x< 1002$
b.  ĐK: $x> \frac{1}{2}$

$\log_2(2x-1)>3$

$\Leftrightarrow 2x-1> 2^3$

$\Leftrightarrow 2x> 9$

$\Leftrightarrow x> \frac{9}{2}$

Vậy $x> \frac{9}{2}$

c. ĐK: $x< -1$

$\log_3(-x-1)\leq 2$

$\Leftrightarrow -x-1\leq 3^2=9$

$\Leftrightarrow x+1\geq -9$

$\Leftrightarrow x\geq -10$

Vậy $-10\leq x< -1$

d. ĐK: $x> \frac{3}{2}$

$\log_2(2x-3)\geq 2$

$\Leftrightarrow 2x-3\geq 2^2=4$

$\Leftrightarrow x\geq \frac{7}{2}$

Vậy $x\geq \frac{7}{2}$

e. ĐK: $x> \frac{7}{2}$

$\log_3(2x-7)>2$
$\Leftrightarrow 2x-7> 3^2=9$
$\Leftrightarrow x> 8$

Vậy $x>8$

1. \(\left|\frac{2x^2-x}{3x-4}\right|\ge1\) Điều kiện: \(x\ne\frac{4}{3}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{2x^2-x}{3x-4}\ge1\\\frac{2x^2-x}{3x-4}\le-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{x^2-2x+2}{3x-4}\ge0\\\frac{x^2+x-2}{3x-4}\le0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x>\frac{4}{3}\\x\in(-\infty;-2]U[1;\frac{4}{3})\end{cases}}\Leftrightarrow x\in(-\infty;-2]U[1;+\infty)\backslash\left\{\frac{4}{3}\right\}\)

2.\(\hept{\begin{cases}x^2\le-2x+3\left(1\right)\\\left(m+1\right)x\ge2m-1\left(2\right)\end{cases}}\)

\(\left(1\right)\Leftrightarrow x^2+2x-3\le0\Leftrightarrow-3\le x\le1\)

+) Nếu \(m=-1\) thì (2) vô nghiệm, suy ra \(m\ne-1\)

+) Nếu \(m>-1\) thì \(\left(2\right)\Leftrightarrow x\ge\frac{2m-1}{m+1}\)

Hệ BPT có nghiệm duy nhất \(\Leftrightarrow\frac{2m-1}{m+1}=1\Leftrightarrow m=2>-1\)

+) Nếu \(m< -1\)thì \(\left(2\right)\Leftrightarrow x\le\frac{2m-1}{m+1}\)

Hệ BPT có nghiệm duy nhất \(\Leftrightarrow\frac{2m-1}{m+1}=-3\Leftrightarrow m=-\frac{2}{5}< -1\)

Vậy \(m=\left\{\frac{-2}{5};2\right\}\)

1. |2x2−x3x−4 |≥1 Điều kiện: x≠43 

⇔[

⇔[

⇔[

⇔x∈(−∞;−2]U[1;+∞)\{43 }

2.{

(1)⇔x2+2x−3≤0⇔−3≤x≤1

∣∣∣∣∣​3x−42x2−x​∣∣∣∣∣​≥1⇔⎣⎢⎢⎢⎡​​3x−42x2−x​≥13x−42x2−x​≤−1​⇔⎣⎢⎢⎢⎡​​3x−42x2−4x+4​≥03x−42x2+2x−4​≤0​⇔⎣⎢⎢⎡​​3x−4>03x−42x2+2x−4​≤0​

\Leftrightarrow \left[\begin{aligned}&{x>\dfrac{4}{3} } \\ &{1\le x<\dfrac{4}{3} } \\ &{x\le -2} \end{aligned}\right.⇔⎣⎢⎢⎢⎢⎢⎡​​x>34​1≤x<34​x≤−2​ .

Tập nghiệm :S=\left(-\infty ;-2\right]\cup \left[1;\dfrac{4}{3} \right)\cup \left(\dfrac{4}{3} ;+\infty \right)S=(−∞;−2]∪[1;34​)∪(34​;+∞).

2.

Ta có: \left\{\begin{aligned}&{x^{2} \le -2x+3} \\ &{\left(m+1\right)x\ge 2m-1} \end{aligned}\right. \Leftrightarrow \left\{\begin{aligned}&{x^{2} +2x-3\le 0} \\ &{\left(m+1\right)x\ge 2m-1} \end{aligned}\right. \Leftrightarrow \left\{\begin{aligned}&{-3\le x\le 1} \\ &{\left(m+1\right)x\ge 2m-1} \end{aligned}\right.{​x2≤−2x+3(m+1)x≥2m−1​⇔{​x2+2x−3≤0(m+1)x≥2m−1​⇔{​−3≤x≤1(m+1)x≥2m−1​.

+ Trường hợp 1: $m=-1$

Hệ BPT trở thành: \left\{\begin{aligned}& {-3\le x\le 1} \\ &{0\ge -3} \end{aligned}\right.{​−3≤x≤10≥−3​. Hệ luôn đúng với $\forall x\in \left[-3;1\right]$.

Vậy $m=-1$ loại.

+ Trường hợp 2: $m>-1$

Hệ BPT trở thành: \left\{\begin{aligned}& {-3\le x\le 1} \\ &{x\ge \dfrac{2m-1}{m+1} } \end{aligned}\right.⎩⎪⎨⎪⎧​​−3≤x≤1x≥m+12m−1​​.

Hệ có nghiệm duy nhất khi \dfrac{2m-1}{m+1} =1\Leftrightarrow 2m-1=m+1\Leftrightarrow m=2m+12m−1​=1⇔2m−1=m+1⇔m=2 (nhận).

+ Trường hợp 3: $m<-1$ Hệ BPT trở thành: \left\{\begin{aligned}& {-3\le x\le 1} \\ &{x\le \dfrac{2m-1}{m+1} } \end{aligned}\right.⎩⎪⎨⎪⎧​​−3≤x≤1x≤m+12m−1​​.

Hệ có nghiệm duy nhất khi \dfrac{2m-1}{m+1} =-3\Leftrightarrow 2m-1=-3m-3\Leftrightarrow m=\dfrac{-2}{5}m+12m−1​=−3⇔2m−1=−3m−3⇔m=5−2​ (loại). Vậy $m=2$ hệ có nghiệm duy nhất.