Bài 2 :

a, \(\frac{1-5x}{x-1}\ge1\)

\(\Leftrightarrow\frac{1-5x}{x-1}\ge\frac{x-1}{x-1}\)

\(\Rightarrow1-5x\ge x-1\)

\(\Leftrightarrow-5x-x\ge-1-1\)

\(\Leftrightarrow-6x\ge-2\)

\(\Leftrightarrow x\le\frac{1}{3}\)

Vậy nghiệm của bất phương trình là \(x\le\frac{1}{3}\).

b, \(\frac{x}{x-2}-\frac{2}{x-3}>1\)

\(\Leftrightarrow\frac{x^2-3x}{x^2-5x+6}-\frac{2x-4}{x^2-5x+6}>\frac{x^2-5x+6}{x^2-5x+6}\)

\(\Rightarrow x^2-3x-2x+4>x^2-5x+6\)

\(\Leftrightarrow x^2-3x-2x-x^2+5x>6-4\)

\(\Leftrightarrow0>2\) ( vô lí )

Vậy bất phương trình vô nghiệm.

Bài 1:

a, \(x^2-8x< 0\)

\(\Leftrightarrow x\left(x-8\right)< 0\)

\(\Leftrightarrow x< 0\) hoặc \(x-8< 0\)

\(\Leftrightarrow x< 0\) hoặc \(x< 8\)

Vậy nghiệm của bất phương trình : x<0 ; x<8

b, \(x^2< 6x-5\)

\(\Leftrightarrow x^2-6x+5< 0\)

\(\Leftrightarrow x^2-x-5x+5< 0\)

\(\Leftrightarrow x\left(x-1\right)-5\left(x-1\right)< 0\)

\(\Leftrightarrow\left(x-1\right)\left(x-5\right)< 0\)

\(\Leftrightarrow\) \(x-1< \) 0 hoặc \(x-5< 0\)

\(\Leftrightarrow x< 1\) hoặc \(x< 5\)

Vậy bất phương trình có nghiệm là x<1 ; x<5

`a)x^2>4`

`<=>sqrtx^2>sqrt4`

`<=>|x|>2`

`<=>` \(\left[ \begin{array}{l}x>2\\x<-2\end{array} \right.\) 

`b)x^2<9`

`<=>\sqrtx^2<sqrt9`

`<=>|x|<3`

`<=>-3<x<3`

`c)(x-1)^2>=4`

`<=>\sqrt{(x-1)^2}>=sqrt4`

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

`<=>` \(\left[ \begin{array}{l}x-1 \ge 2\\x-1 \le -2\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x \ge 3\\x \le -1\end{array} \right.\) 

`d)(1-2x)^2<=0,09`

`<=>\sqrt{(1-2x)^2}<=sqrt{0,09}`

`<=>|2x-1|<=0,3`

`<=>-0,3<=2x-1<=0,3`

`<=>0,7<=2x<=1,3`

`<=>0,35<=x<=0,65`

`e)x^2+6x-7>0`

`<=>x^2-x+7x-7>0`

`<=>x(x-1)+7(x-1)>0`

`<=>(x-1)(x+7)>0`

TH1:

\(\left[ \begin{array}{l}x-1>0\\x+7>0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x>1\\x>-7\end{array} \right.\) 

`<=>x>1`

TH2"

\(\left[ \begin{array}{l}x-1<0\\x+7<0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x<1\\x<-7\end{array} \right.\) 

`<=>x<-7`

`f)x^2-x<2`

`<=>x^2-x-2<0`

`<=>x^2-2x+x-2<0`

`<=>x(x-2)+x-2<0`

`<=>(x-2)(x+1)<0`

`<=>` \(\begin{cases}x-2<0\\x+1>0\\\end{cases}\)

`<=>` \(\begin{cases}x<2\\x>-1\\\end{cases}\)

`<=>-1<x<2`

a) x² > 4

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

b) \(x^2< 9\)

<=> \(-3< x< 3\)

c) \(\left(x-1\right)^2\ge4\)

<=> \(\left[{}\begin{matrix}x-1\ge2< =>x\ge3\\x-1\le-2< =>x\le-1\end{matrix}\right.\)

d) \(\left(1-2x\right)^2\le0,09\)

<=> \(-0,3\le1-2x\le0,3\)

<=> \(1,3\ge2x\ge0,7\)

<=> \(0,65\ge x\ge0,35\)

e) \(x^2+6x-7>0\)

<=> \(\left(x+7\right)\left(x-1\right)>0\)

<=> \(\left[{}\begin{matrix}x-1>0< =>x>1\\x+7< 0< =>x< -7\end{matrix}\right.\)

f) \(x^2-x< 2\)

<=> \(x^2-x-2< 0\)

<=> \(\left(x-2\right)\left(x+1\right)< 0\)

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

<=> -1 < x < 2

g) \(4x^2-12x\le\dfrac{-135}{16}\)

<=> \(64x^2-192x+135\le0\)

<=> (8x - 15)(8x - 9) \(\le0\)

<=> \(\left\{{}\begin{matrix}8x-15\le0< =>x\le\dfrac{15}{8}\\8x-9\ge0< =>x\ge\dfrac{9}{8}\end{matrix}\right.\)

<=> \(\dfrac{9}{8}\le x\le\dfrac{15}{8}\)

a, \(\dfrac{x+1}{x+3}>1\Leftrightarrow\dfrac{x+1}{x+3}-1>0\Leftrightarrow\dfrac{x+1-x-3}{x+3}>0\)

\(\Rightarrow x+3< 0\)do  -2 < 0 

\(\Rightarrow x< -3\)Vậy tập nghiệm BFT là S = { x | x < -3 } 

b, \(\dfrac{2x-1}{x-3}\le2\Leftrightarrow\dfrac{2x-1}{x-3}-2\le0\Leftrightarrow\dfrac{2x-1-2x+6}{x-3}\le0\)

\(\Rightarrow x-3\le0\)do 5 > 0 

\(\Rightarrow x\le3\)Vậy tập nghiệm BFT là S = { x | x \(\le\)3 } 

c, \(\dfrac{x^2+2x+2}{x^2+3}\ge1\Leftrightarrow\dfrac{x^2+2x+2}{x^2+3}-1\ge0\)

\(\Leftrightarrow\dfrac{x^2+2x+2-x^2-3}{x^2+3}\ge0\Rightarrow2x-1\ge0\)do x^2 + 3 > 0 

\(\Rightarrow x\ge\dfrac{1}{2}\)Vậy tập nghiệm BFT là S = { x | x \(\ge\)1/2 } 

mình ko chắc nên mình đăng sau :> 

d, \(\dfrac{2x+1}{x^2+2}\ge1\Leftrightarrow\dfrac{2x+1}{x^2+2}-1\ge0\Leftrightarrow\dfrac{2x+1-x^2-2}{x^2+2}\ge0\)

\(\Rightarrow-x^2+2x-1\ge0\Rightarrow-\left(x-1\right)^2\ge0\)vô lí 

\(\dfrac{x-3}{2+x}< 1\\ \Leftrightarrow\dfrac{x-3}{2+x}-1< 0\\ \Leftrightarrow\dfrac{x-3}{2+x}-\dfrac{2+x}{2+x}< 0\\ \Leftrightarrow\dfrac{x-3-2-x}{2+x}< 0\\ \Leftrightarrow\dfrac{-5}{2+x}< 0\)

Vì \(-5< 0\)

\(\Rightarrow2+x>0\\ \Rightarrow x>-2\)

Vậy \(x>-2\)

Bài 1:

a. 

$(4x^2+4x+1)-x^2=0$

$\Leftrightarrow (2x+1)^2-x^2=0$

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

$\Leftrightarrow (x+1)(3x+1)=0$

$\Rightarrow x+1=0$ hoặc $3x+1=0$

$\Rightarrow x=-1$ hoặc $x=-\frac{1}{3}$

b.

$x^2-2x+1=4$

$\Leftrightarrow (x-1)^2=2^2$

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

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

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

$\Leftrightarrow x-3=0$ hoặc $x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-1$

c.

$x^2-5x+6=0$

$\Leftrightarrow (x^2-2x)-(3x-6)=0$

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

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

$\Leftrightarrow x-2=0$ hoặc $x-3=0$

$\Leftrightarrow x=2$ hoặc $x=3$

2c.

ĐKXĐ: $x\neq 0$

PT $\Leftrightarrow x-\frac{6}{x}=x+\frac{3}{2}$

$\Leftrightarrow -\frac{6}{x}=\frac{3}{2}$

$\Leftrightarrow x=-4$ (tm)

2d.

ĐKXĐ: $x\neq 2$

PT $\Leftrightarrow \frac{1+3(x-2)}{x-2}=\frac{3-x}{x-2}$

$\Leftrightarrow \frac{3x-5}{x-2}=\frac{3-x}{x-2}$

$\Rightarrow 3x-5=3-x$

$\Leftrightarrow 4x=8$

$\Leftrightarrow x=2$ (không tm) 

Vậy pt vô nghiệm.

2f.

ĐKXĐ: $x\neq \pm 2$

PT $\Leftrightarrow \frac{(x-2)^2-3(x+2)}{(x+2)(x-2)}=\frac{2(x-11)}{(x-2)(x+2)}$

$\Rightarrow (x-2)^2-3(x+2)=2(x-11)$

$\Leftrightarrow x^2-4x+4-3x-6=2x-22$

$\Leftrightarrow x^2-7x-2=2x-22$

$\Leftrightarrow x^2-9x+20=0$

$\Leftrightarrow (x-4)(x-5)=0$

$\Leftrightarrow x-4=0$ hoặc $x-5=0$

$\Leftrightarrow x=4$ hoặc $x=5$ (tm)

\(a,PT\Leftrightarrow\left|3x-1\right|=\left|x-3\right|\Leftrightarrow\left[{}\begin{matrix}3x-1=x-3\\3x-1=3-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\\ b,PT\Leftrightarrow\left|x-4\right|=4-x\Leftrightarrow\left[{}\begin{matrix}x-4=4-x\left(x\ge4\right)\\x-4=x-4\left(x< 4\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)