9) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)

\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)

\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)

\(\Leftrightarrow-4x=9\)

hay \(x=-\dfrac{9}{4}\)

10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}

11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)

\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{9x+5}{\left(x-4\right)\left(x+5\right)}\)

Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)

\(\Leftrightarrow5x^2-7x=0\)

\(\Leftrightarrow x\left(5x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)

12) Ta có: \(\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{\left(2x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow\dfrac{2x\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{4x^2-1+4}{\left(2x-1\right)\left(2x+1\right)}\)

Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)

\(\Leftrightarrow2x^2+x-3=0\)

\(\Leftrightarrow2x^2+3x-2x-3=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)=0\)

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

13) Ta có: \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\)

\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)

Suy ra: \(x^2+2x-x+2-2=0\)

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

a)  3x – 15 = 25 – 5x 

=> 3x + 5x = 25 + 15

=> 8x = 40

=> x = 5

 b) 3x - 17 = 2x – 7     

=> 3x - 2x = -7 + 17

=> x = 10

 c) 2x – 17 =  – (3x – 18)

=> 2x - 17 = -3x + 18

=> 2x + 3x = 18 + 17

=> 5x = 35

=> x = 7

d) 3x – 14 = 2(x – 9) + 1

=> 3x - 14 = 2x - 18 + 1

=> 3x - 2x = -18 + 1 + 14

=> x = -3

f) (x – 5)² = 9          

\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

a) Ta có: \(3x-15=25-5x\)

\(\Leftrightarrow3x-15-25+5x=0\)

\(\Leftrightarrow8x-40=0\)

\(\Leftrightarrow8x=40\)

hay x=5

Vậy: x=5

b) Ta có: \(3x-17=2x-7\)

\(\Leftrightarrow3x-17-2x+7=0\)

\(\Leftrightarrow x-10=0\)

hay x=10

Vậy: x=10

c) Ta có: \(2x-17=-\left(3x-18\right)\)

\(\Leftrightarrow2x-17=-3x+18\)

\(\Leftrightarrow2x-17+3x-18=0\)

\(\Leftrightarrow5x-35=0\)

\(\Leftrightarrow5x=35\)

hay x=7

Vậy: x=7

d) Ta có: \(3x-14=2\left(x-9\right)+1\)

\(\Leftrightarrow3x-14=2x-18+1\)

\(\Leftrightarrow3x-14-2x+18-1=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Vậy: x=-3

f) Ta có: \(\left(x-5\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

Vậy: \(x\in\left\{2;8\right\}\)

Bài 2. Tìm x, biết :                                

a) \(3x-15=25-5x\)

\(\Leftrightarrow8x=40\)

\(\Leftrightarrow x=5\)

Vậy x = 5

b) \(3x-17=2x-7\)

\(\Leftrightarrow x=10\)

Vậy x = 10

c) \(2x-17=-\left(3x-18\right)\)

\(\Leftrightarrow2x-17=18-3x\)

\(\Leftrightarrow5x=35\)

\(\Leftrightarrow x=7\)

Vậy x = 7

d) \(3x-14=2\left(x-9\right)+1\)

\(\Leftrightarrow3x-14=2x-18+1\)

\(\Leftrightarrow3x-14=2x-17\)

\(\Leftrightarrow x=-3\)

Vậy x = -3

e) \(\left(x-5\right)^2=9\)

\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

Vậy x = {8; 2}

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14) Ta có: \(\dfrac{x}{2x-6}+\dfrac{x}{2x+2}=\dfrac{2x+4}{x^2-2x-3}\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{4x+8}{2\left(x-3\right)\left(x+1\right)}\)

Suy ra: \(x^2+x+x^2-3x-4x-8=0\)

\(\Leftrightarrow2x^2-6x-8=0\)

\(\Leftrightarrow x^2-3x-4=0\)

a=1; b=-3; c=-4

Vì a-b+c=0 nên phương trình có hai nghiệm phân biệt là:

\(x_1=-1\left(loại\right);x_2=\dfrac{-c}{a}=4\left(nhận\right)\)

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

3x+5 =0 hoặc 4-3x=0

3x=-5 hoặc 3x=-4

x=-5/3 hoặc x=-4/3

9(3x-2)=x(2-3x)

9(3x-2)-x(3x-2)=0

(3x-2)(9-x)=0

3x-2=0 hoặc 9-x=0

3x=2 hoặc x= -9

x =2/3 hoặc x=-9 

vậy x =2/3 ; x= -9

25x^2 - 2=0

(5x)^2 -√2^2=0

(5x-√2)(5x+√2)=0

5x=√2 hoặc 5x = -√2

x=√2/5 hoặc x= -√2/5

vậy x=√2/5 ; x=-√2/5

b, x = -5/3 hoặc x = 4/3.

c, x = 0 hoặc x = 3, -3.

d, x = 0 hoặc x = 2, -2.

e, x = 1 hoặc x = \(\dfrac{-1}{2}\).

a: \(\Leftrightarrow x^2-40x+400-x^2-4x-3=-7\)

=>-44x+397=-7

=>-44x=-404

hay x=101

b: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=0\\4-3x=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{5}{3};\dfrac{4}{3}\right\}\)

c: \(\Leftrightarrow x\left(x^2-9\right)=0\)

=>x(x-3)(x+3)=0

hay \(x\in\left\{0;3;-3\right\}\)

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

hay \(x\in\left\{0;2;-2\right\}\)

e: =>(2x+1)(1-x)=0

=>x=-1/2 hoặc x=1

+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0

-2x^4 + 3x^3-2x^2=0

x^2(-2x^2+x-2)=0

-2x^2(x-1)^2=0

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2