1:

a: =>3x=6

=>x=2

b: =>4x=16

=>x=4

c: =>4x-6=9-x

=>5x=15

=>x=3

d: =>7x-12=x+6

=>6x=18

=>x=3

2:

a: =>2x<=-8

=>x<=-4

b: =>x+5<0

=>x<-5

c: =>2x>8

=>x>4

ĐKXĐ: \(2\le x\le5\)

\(\left(\sqrt{2x-4}-\sqrt{5-x}\right)\sqrt{3x-3}=3x-9\)

\(\Leftrightarrow\dfrac{\left(3x-9\right)\sqrt{3x-3}}{\sqrt{2x-4}+\sqrt{5-x}}=3x-9\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-9=0\Rightarrow x=3\\\dfrac{\sqrt{3x-3}}{\sqrt{2x-4}+\sqrt{5-x}}=1\left(1\right)\end{matrix}\right.\)

Xét (1):

\(\Leftrightarrow\sqrt{3x-3}=\sqrt{2x-4}+\sqrt{5-x}\)

\(\Leftrightarrow3x-3=x+1+2\sqrt{\left(2x-4\right)\left(5-x\right)}\)

\(\Leftrightarrow x-2=\sqrt{\left(2x-4\right)\left(5-x\right)}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left(x-2\right)^2=\left(2x-4\right)\left(5-x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left(x-2\right)\left(3x-12\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

Vậy pt có 3 nghiệm \(x=\left\{2;3;4\right\}\)

a. 3(x-2)-10=5(2x + 1)

<=> 3x - 6 - 10 = 10x + 5

<=> 3x - 10x = 5 + 6 + 10

<=> -7x = 21

<=> x = -3

b. 3x + 2=8 -2(x-7)

<=> 3x + 2 = 8 - 2x + 14

<=> 3x + 2x = 8 + 14 - 2

<=> 5x = 20

<=> x = 4

c. 2x-(2+5x)= 4(x + 3)

<=> 2x - 2 - 5x = 4x + 12

<=> 2x - 5x - 4x = 12 + 2

<=> -7x = 14

<=> x = -2

d. 5-(x +8)=3x + 3(x-9)

<=> 5 - x - 8 = 3x + 3x - 27

<=> -x - 3x - 3x = -27 + 8 - 5

<=> -7x = -24

<=> x = 24/7

e. 3x - 18 + x= 12-(5x + 3)

<=> 3x - 18 + x = 12 - 5x - 3

<=> 3x + x - 5x = 12 - 3 + 18

<=> -x = 27

<=> x = - 27

a. 3(x-2)-10=5(2x + 1)

<=> 3x - 6 - 10 = 10x + 5

<=> 3x - 10x = 5 + 6 + 10

<=> -7x = 21

<=> x = -3

b. 3x + 2=8 -2(x-7)

<=> 3x + 2 = 8 - 2x + 14

<=> 3x + 2x = 8 + 14 - 2

<=> 5x = 20

<=> x = 4

c. 2x-(2+5x)= 4(x + 3)

<=> 2x - 2 - 5x = 4x + 12

<=> 2x - 5x - 4x = 12 + 2

<=> -7x = 14

<=> x = -2

d. 5-(x +8)=3x + 3(x-9)

<=> 5 - x - 8 = 3x + 3x - 27

<=> -x - 3x - 3x = -27 + 8 - 5

<=> -7x = -24

<=> x = 24/7

e. 3x - 18 + x= 12-(5x + 3)

<=> 3x - 18 + x = 12 - 5x - 3

<=> 3x + x - 5x = 12 - 3 + 18

<=> -x = 27

<=> x = - 27

\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)

\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)

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

\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)

\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)

\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

3.15:

a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)

b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

c, \(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

3.16

\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)

\(\Leftrightarrow-14m+35-2m^2+8=0\)

\(\Leftrightarrow-14m-2m^2+43=0\)

\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)

\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)

\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)

\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)

pt vô nghiệm

a) 3x-6=0

3x=6 => x=2

b) (3x+2)(4x-5)=0

=> 3x+2=0 => x=-2/3

hoặc 4x-5=0 => x=5/4

câu c ,d thiếu dấu '=" để thành 1 pt rồi bạn

c) \(\dfrac{2x-5}{3}+\dfrac{x-3}{5}=\dfrac{4x+3}{15}\)

=> 10x -25 +3X-9=4X+3

=>9x=37

=>x=37/9

d) \(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-5}{x^2-9}\) ĐK (x khác 3,-3)

=>5x+15+4x-12=x-5

=>8x=-8

=>x=-1

đề câu 2 có sai gì ko v 

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)