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 $x>8$

a: \(log\left(x-5\right)< 2\)

=>\(\left\{{}\begin{matrix}x-5>0\\log\left(x-5\right)< log4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-5>0\\x-5< 4\end{matrix}\right.\Leftrightarrow5< x< 9\)

b: \(log_2\left(2x-3\right)>4\)

=>\(log_2\left(2x-3\right)>log_216\)

=>\(\left\{{}\begin{matrix}2x-3>0\\2x-3>16\end{matrix}\right.\)

=>2x-3>16

=>2x>19

=>\(x>\dfrac{19}{2}\)

c: \(log_3\left(2x+5\right)< =3\)

=>\(log_3\left(2x+5\right)< =log_327\)

=>\(\left\{{}\begin{matrix}2x+5>0\\2x+5< =27\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>-\dfrac{5}{2}\\x< =11\end{matrix}\right.\)

=>\(-\dfrac{5}{2}< x< =11\)

d: \(log_4\left(4x-5\right)>=2\)

=>\(log_4\left(4x-5\right)>=log_416\)

=>4x-5>=16 và 4x-5>0

=>4x>=21 và 4x>5

=>4x>=21

=>\(x>=\dfrac{21}{4}\)

e: \(log_3\left(1-3x\right)>3\)

=>\(log_3\left(1-3x\right)>log_327\)

=>\(\left\{{}\begin{matrix}1-3x>0\\1-3x>27\end{matrix}\right.\)

=>1-3x>27

=>\(-3x>26\)

=>\(x< -\dfrac{26}{3}\)

a,Áp dụng BĐT `|A|-|B|<=|A-B|`

`=>|x+1|-|x-2|<=|x+1-x+2|=3`

Mà đề bài `|x+1|-|x-2|>=3`

`=>|x+1|-|x-2|=3`

`=>x=2\or\x=-1`

`b,1/(|x|-3)-1/2<0`

`<=>(5-|x|)/(2|x|-6)<0`

`<=>(|x|-5)/(|x|-3)>0`

`<=>` $\left[ \begin{array}{l}|x|>5\\|x|<3\end{array} \right.$

`<=>` $\left[ \begin{array}{l}\left[ \begin{array}{l}x>5\\x<-5\end{array} \right.\\-3<x<3\end{array} \right.$

`-2<=x<=1` nhé câu a ý mình ghi thiếu.

a, \(\dfrac{\left(2x-5\right)\left(x+2\right)}{4x-3}< 0\)

⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}\left(2x-5\right)\left(x+2\right)< 0\\4x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}\left(2x-5\right)\left(x+2\right)>0\\4x-3< 0\end{matrix}\right.\end{matrix}\right.\)

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

⇔ \(\left[{}\begin{matrix}\dfrac{3}{4}< x< \dfrac{5}{2}\\x< -2\end{matrix}\right.\)

Vậy tập nghiệm của bất phương trình là

S = \(\left(\dfrac{3}{4};\dfrac{5}{2}\right)\cup\left(-\infty;-2\right)\)

b, Pt

⇔ \(\left\{{}\begin{matrix}x^2-5x+6=x^2+6x+5\\x\in R\backslash\left\{-1;2\right\}\end{matrix}\right.\)

⇔ x = \(\dfrac{1}{11}\)

Vậy S = \(\left\{\dfrac{1}{11}\right\}\)

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

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                                                        ⇔$-{}^{}$

Lời giải:

ĐK: $x\neq -5; n\neq 0$

\(\frac{(2x+1)^4(x-3)^3}{(x+5)^2x^5}\leq 0\Leftrightarrow \left[\frac{(2x+1)^2(x-3)}{(x+5)x^2}\right]^2.\frac{x-3}{x}\leq 0\)

\(\Leftrightarrow \frac{x-3}{x}\leq 0\Rightarrow \left[\begin{matrix} x-3\geq 0; x< 0\\ x-3\leq 0; x>0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} 0> x\geq 3(\text{vô lý})\\ 3\geq x>0\end{matrix}\right.\)

Vậy $3\geq x>0$

Đk: \(x\ge\dfrac{1}{2}\)

Bpt\(\Leftrightarrow\left(x^2+2x\sqrt{2x-1}+2x-1\right)-\left[4\left(2x-1\right)+4\sqrt{2x-1}+1\right]\ge0\)

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

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

Vì \(x\ge\dfrac{1}{2}\Rightarrow x+3\sqrt{2x-1}+1>0\)

Từ (1) \(\Rightarrow x-\sqrt{2x-1}-1\ge0\)

\(\Leftrightarrow\sqrt{2x-1}\le x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\ge0\\x-1\ge0\\2x-1\le\left(1-x\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\in R\backslash\left(2-\sqrt{2};2+\sqrt{2}\right)\end{matrix}\right.\)\(\Rightarrow x\ge2+\sqrt{2}\)

Vậy...

\(\dfrac{15x-2}{4}-\dfrac{x^2+1}{3}>\dfrac{x\left(1-2x\right)}{6}+\dfrac{x-3}{2}\\ \Leftrightarrow3\left(15x-2\right)-4\left(x^2+1\right)>2x\left(1-2x\right)+6\left(x-3\right)\\ \Leftrightarrow45x-6-4x^2-4>2x-4x^2+6x-18\\ \Leftrightarrow45x-6x-2x>6+4-18\\ \Leftrightarrow37x>-8\\ \Leftrightarrow x>-\dfrac{8}{37}\)

\(\dfrac{3\left(15x-2\right)}{12}-\dfrac{4\left(x^2+1\right)}{12}>\dfrac{2x\left(1-2x\right)}{12}+\dfrac{6\left(x-3\right)}{12}\)

\(45x-6-\left(4x^2+4\right)>2x-4x^2+6x-18\)

\(45x-4x^2+4x^2-2x-6x>6+4-18\)

\(37x>-8\)

\(x>\dfrac{-8}{37}\)