\(\dfrac{3}{2x+10}+\dfrac{3}{x-5}-\dfrac{2x}{x^2-25}\)

\(=\dfrac{3\left(x-5\right)}{2\left(x+5\right)\left(x-5\right)}+\dfrac{6\left(x+5\right)}{2\left(x+5\right)\left(x-5\right)}-\dfrac{4x}{2\left(x+5\right)\left(x-5\right)}\)

\(=\dfrac{3x-15+6x+30-4x}{2\left(x+5\right)\left(x-5\right)}\)

\(=\dfrac{5x+15}{2\left(x+5\right)\left(x-5\right)}\)

`@` `\text {Ans}`

`\downarrow`

`a,`

`(2x - 3)^2`

`= 4x^2 - 12x + 9`

`b,`

`(x + 1)^2`

`= x^2 + 2x + 1`

`c,`

`(2x + 5)(2x - 5)`

`= 4x^2 - 25`

`d,`

`(a + b - c)(a - b + c)`

`= a^2 - b^2 + bc - c^2 + cb`

`e,`

\((x + 1)^2 - 10(x + 1) + 25\)

`= x^2 + 2x + 1 - 10x - 10 + 25`

`= x^2 - 8x +16`

`@` `\text {Kaizuu lv uuu}`

`@` CT:

Bình phương của `1` tổng: `(A + B)^2 = A^2 + 2AB + B^2`

Bình phương của `1` hiệu: `(A - B)^2 = A^2 - 2AB + B^2`

`A^2 - B^2 = (A-B)(A+B)`

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: =>5-x=-23

=>x=5+23=28

b: =>x-3-x+7-25+x=54

=>x-21=54

=>x=75

c: =>7-9x-2x+4=-5x-35+27-25=-5x-37

=>-11x+3=-5x-37

=>-6x=-40

=>x=20/3

a) \(10-x-5=\left(-5\right)-7-11\)

\(\Rightarrow5-x=-23\)

\(\Rightarrow x=5+23\)

\(\Rightarrow x=28\)

b) \(\left(x-3\right)-\left(x+17-24\right)-\left(25-x\right)=24-\left(-30\right)\)

\(\Rightarrow x-3-x-17+24-25+x=54\)

\(\Rightarrow x-21=54\)

\(\Rightarrow x=54+21\)

\(\Rightarrow x=75\)

c) \(\left(7-9x\right)-\left(2x-4\right)=-\left(5x+35\right)-\left(-27\right)-25\)

\(\Rightarrow7-9x-2x+4=-5x-35+27-25\)

\(\Rightarrow11-11x=-5x-33\)

\(\Rightarrow-11x+5x=-33-11\)

\(\Rightarrow-6x=-22\)

\(\Rightarrow x=\dfrac{22}{6}=\dfrac{11}{3}\)

a. 
10-x-5 = (-5) - 7 -11
=>5-x = 0
=>x=5
b
(x-3) - (x+17-24) - (25-x) = 24 - (-30)
=>x - 3 - x - 17 + 24 - 25 - x = 24 + 30
=>-x - 21 = 54
=>-x = 75
=>x = -75
c
(7 - 9x) - (2x - 4) = - (5x + 35) - (-27) - 25
=>7-9x - 2x + 4 = -5x - 35 + 27 - 35
=>11 - 11x + 5x = -43
=>16x = 11 + 43
=>16x = 54
=>x=4

a, 2.x + 7 = 15

   2x         = 8

   x           = 4

b, 25 – 3.(6 – x) = 22

            3.(6-x)    = 3

               6-x      = 1

                  x      = 5

c, [(2x – 11) : 3 + 1].5 = 20

      (2x-11) : 3+1         = 4

       (2x-11):3              = 3

        2x-11                  = 1

        2x                       = 12

        x                         = 6

  e, 2 . 3x = 10 . 312 + 8 . 274

      6x      = 3120 + 2192

      6x      = 5312

        x      = 5312/6

g, x – 12 = (–8) + (–17)

    x  - 12 = -25

    x         = -13

Lần sau tách nhỏ nội dung câu hỏi ra nha em, chứ trả lời thế này biếng lắm '^^ Chị làm chỉ mang tính tham khảo kết quả thôi, còn cụ thể thì em tách từng bước một ra he :>

a, \(2\cdot x+7=15\)

\(\Leftrightarrow2\cdot x=8\)

\(\Leftrightarrow x=4\)

Vậy x = 4.

b, \(25-3\cdot\left(6-x\right)=22\)

\(\Leftrightarrow3\cdot\left(6-x\right)=3\)

\(\Leftrightarrow6-x=1\)

\(\Leftrightarrow x=5\)

Vậy x = 5.

c, \(\left[\left(2x-11\right):3+1\right]\cdot5=20\)

\(\Leftrightarrow\left(2x-11\right):3+1=4\)

\(\Leftrightarrow\left(2x-11\right):3=3\)

\(\Leftrightarrow2x-11=9\)

\(\Leftrightarrow2x=20\)

\(\Leftrightarrow x=10\)

Vậy x = 10.

d, \(\left(25-2x\right)\cdot3:5-32=42\)

\(\Leftrightarrow\)\(\frac{3\cdot\left(25-2x\right)}{5}=74\)

\(\Leftrightarrow3\cdot\left(25-2x\right)=370\)

\(\Leftrightarrow25-2x=\frac{370}{3}\)

\(\Leftrightarrow2x=-\frac{295}{3}\)

\(\Leftrightarrow x\approx49\)

Vậy \(x\approx49\) .

e, \(2\cdot3x=10\cdot312+8\cdot274\)

\(\Leftrightarrow6x=5312\)

\(\Leftrightarrow x=5312:6\approx885\)

Vậy \(x\approx885\) .

g, \(x-12=\left(-8\right)+\left(-17\right)\)

\(\Leftrightarrow x-12=-25\)

\(\Leftrightarrow x=-25+12=-13\)

Vậy x = -13.

h, \(7-2x=18-3x\)

\(\Leftrightarrow-2x+3x=18-7\)

\(\Leftrightarrow x=11\)

Vậy \(x=11\) .

i, \(3\cdot\left(x+5\right)-x-11=24\)

\(\Leftrightarrow3x+15-x-11=24\)

\(\Leftrightarrow2x=24+11-15\)

\(\Leftrightarrow2x=20\)

\(\Leftrightarrow x=10\)

Vậy \(x=10\) .

Thanks mn ^v^

g. \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)

Vậy \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)

f. \(\frac{2}{3}x-\frac{1}{2}x=\frac{5}{12}\)

\(\Leftrightarrow x\left(\frac{2}{3}-\frac{1}{2}\right)=\frac{5}{12}\)

\(\Leftrightarrow x\left(\frac{4}{6}-\frac{3}{6}\right)=\frac{5}{12}\)

\(\Leftrightarrow\frac{1}{6}x=\frac{5}{12}\)

\(\Leftrightarrow x=\frac{5}{12}\div\frac{1}{6}\)

\(\Leftrightarrow x=\frac{30}{12}=\frac{5}{2}\)

c. \(3x+2x=\frac{5}{8}\times\frac{4}{3}\)

\(\Leftrightarrow x\left(3+2\right)=\frac{5.4}{8.3}\)

\(\Leftrightarrow5x=\frac{5.1}{2.3}\)

\(\Leftrightarrow5x=\frac{5}{6}\)

\(\Leftrightarrow x=\frac{5}{6}\div5\)

\(\Leftrightarrow x=\frac{1}{6}\)

a: =>10căn x=30

=>căn x=3

=>x=9

c: =>|1/2x-3/4|=1/2

=>1/2x-3/4=1/2 hoặc 1/2x-3/4=-1/2

=>1/2x=5/4 hoặc 1/2x=1/2

=>x=1 hoặc x=5/2

3) \(-12+2x-9+x=0\\ -21+3x=0\\ 3x=21\\ x=7\)

4)
\(11+\left(15-x\right)=1\)
\(15-x=1-11\)
\(15-x=-10\)
\(x=15-\left(-10\right)\)
\(x=25\)

5) 
\(4-\left(27-3\right)=x-\left(13-4\right)\)
\(4-24=x-9\)
\(x-9=-20\)
\(x=-20+9\)
\(x=-11\)
 

\(3.-12+\left(2x-9\right)+x=0.\)

\(\Leftrightarrow-12+2x-9+x=0.\Leftrightarrow3x=21.\Leftrightarrow x=7.\)

Vậy \(x=7.\)

\(4.11+\left(15-x\right)=1.\Leftrightarrow11+15-x=1.\Leftrightarrow26-x=1.\Leftrightarrow x=25.\)

Vậy \(x=25.\)

\(5.4-\left(27-3\right)=x-\left(13-4\right).\Leftrightarrow4-24=x-9.\Leftrightarrow-20=x-9.\Leftrightarrow x=-11.\)

Vậy \(x=-11.\)

\(6.8-\left(x-10\right)=23-\left(-4+12\right).\Leftrightarrow8-x+10=23-8.\Leftrightarrow18-x=15.\Leftrightarrow x=3.\)

Vậy \(x=3.\)

\(7.105-5\left(10-5x\right)=-20.\Leftrightarrow105-50+25x=-20.\Leftrightarrow25x=-75.\Leftrightarrow x=-3.\)

Vậy \(x=-3.\)

\(8.\left(x-1\right)\left(8-2x\right)\left(3x+123\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\8-2x=0.\\3x+123=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=4.\\x=-41.\end{matrix}\right.\)

Vậy \(x\in\left\{1;4;-41\right\}.\)

\(9.\left(x^2-25\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0.\\x+5=0.\\x+10=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\\x=-10.\end{matrix}\right.\)

Vậy \(x\in\left\{5;-5;-10\right\}.\)

\(10.x\left(x^2+5\right)=0.\Leftrightarrow x=0.\)