Câu 1: 

\(\Rightarrow \left[\begin{array}{} x+\frac{1}{2}=0\\ \frac{2}{3}-2x=0 \end{array} \right.\)

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

Vậy phương trình có tập nghiệm S={\(\frac{-1}{2};\frac{1}{3}\)}

Câu 2: 

\(\Rightarrow \left[\begin{array}{} 3x-10=0\\ 5-\frac{1}{2}x=0 \end{array} \right.\)

\(\Leftrightarrow \left[\begin{array}{} x-=\frac{10}{3}\\ x=10 \end{array} \right.\)

Vậy phương trình có tập nghiệm S={\(10;\frac{10}{3}\)}

Câu 3: 

\(\Leftrightarrow \frac{1}{3}x=\frac{65}{4}-\frac{53}{4}\)

\( \Leftrightarrow \frac{1}{3}x=\frac{12}{4}\)

\(\Leftrightarrow x=9\)

Vậy phương trình có tập nghiệm S={9}

Câu 4: 

\(\Leftrightarrow \frac{2}{3}x=\frac{2}{3}\)

\(\Leftrightarrow x=1\)

Vậy phương trình có tập nghiệm S={1}

Câu 5: 

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x(x+1)}=\frac{2010}{2011}\)

\(\Leftrightarrow 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2010}{2011}\)

\(\Leftrightarrow 1-\frac{1}{x+1}=\frac{2010}{2011}\)

\(\Leftrightarrow \frac{x}{x+1}=\frac{2010}{2011}\)

\(\Rightarrow 2010x+2010=2011x\)

\(\Leftrightarrow x=2010\)

Vậy phương trình có tập nghiệm S={2010}

cảm ơn bạn Hoàng Bình Bảo nha nhưng mà đây là toán lớp 6 mà bạn 

Giải:

1)

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)

\(\dfrac{6}{5}-x=\dfrac{2}{3}\)

\(x=\dfrac{6}{5}-\dfrac{2}{3}\)

\(x=\dfrac{18}{15}-\dfrac{10}{15}\)

\(x=\dfrac{8}{15}\)

Vậy, `x =`\(\dfrac{8}{15}\)

`b)`

\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)

\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)

\(x=\dfrac{4}{17}\)

Vậy, \(x=\dfrac{4}{17}\)

`c)`

\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)

\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)

\(x=\dfrac{34}{7}\)

Vậy, `x = `\(\dfrac{34}{7}\)

a) \(\dfrac{3}{2}x\dfrac{4}{5}-x=\dfrac{2}{3}\Rightarrow\dfrac{6}{5}-x=\dfrac{2}{3}\Rightarrow x=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{18}{15}-\dfrac{10}{15}=\dfrac{8}{15}\)

b) \(x.3\dfrac{1}{3}=3\dfrac{1}{3}:4\dfrac{1}{4}\Rightarrow\dfrac{10}{3}.x=\dfrac{10}{3}:\dfrac{17}{4}\Rightarrow\dfrac{10}{3}.x=\dfrac{10}{3}.\dfrac{4}{17}\Rightarrow x=\dfrac{10}{3}.\dfrac{4}{17}:\dfrac{10}{3}=\dfrac{10}{3}.\dfrac{4}{17}.\dfrac{3}{10}=\dfrac{4}{17}\)

c) \(5\dfrac{2}{3}:x=3\dfrac{2}{3}-2\dfrac{1}{2}\Rightarrow\dfrac{17}{3}:x=\dfrac{11}{3}-\dfrac{5}{2}\Rightarrow\dfrac{17}{3}:x=\dfrac{22}{6}-\dfrac{15}{6}\Rightarrow\dfrac{17}{3}:x=\dfrac{7}{6}\Rightarrow x=\dfrac{17}{3}:\dfrac{7}{6}=\dfrac{17}{3}.\dfrac{7}{6}=\dfrac{119}{18}\)

a, \(\dfrac{x-1}{21}\) = \(\dfrac{3}{x+1}\)

   ( x-1)(x+1) = 21.3

    x² + x - x -1 = 63

     x²                = 63 + 1

     x²               = 64

    x = + - 8

b, 2\(\dfrac{1}{2}\)x + x = 2\(\dfrac{1}{17}\)

        x( \(\dfrac{5}{2}\) + 1) = \(\dfrac{35}{17}\)

       x              = \(\dfrac{35}{17}\) : ( \(\dfrac{5}{2}\)+1)

       x             = \(\dfrac{35}{17}\) x \(\dfrac{2}{7}\)

       x            = \(\dfrac{10}{17}\)

c, (x + \(\dfrac{1}{4}\) - \(\dfrac{2}{3}\) ) : ( 2 + \(\dfrac{1}{6}\) - \(\dfrac{1}{4}\)) = \(\dfrac{7}{46}\)

   (x  - \(\dfrac{5}{12}\)):  \(\dfrac{23}{12}\)                     =   \(\dfrac{7}{46}\)

  (x - \(\dfrac{5}{12}\))                               =   \(\dfrac{7}{46}\) x \(\dfrac{23}{12}\)

  x   - \(\dfrac{5}{12}\)                                =    \(\dfrac{7}{12}\)

 x                                            =    \(\dfrac{7}{12}\) + \(\dfrac{5}{12}\)

x                                             =     1

d, 2\(\dfrac{1}{3}\)x - 1\(\dfrac{3}{4}\)x + \(2\dfrac{2}{3}\)  = 3\(\dfrac{3}{5}\)

   x( \(\dfrac{7}{3}\) - \(\dfrac{7}{4}\)) + \(\dfrac{8}{3}\)      =  \(\dfrac{18}{5}\)

   x\(\dfrac{7}{12}\)                    = \(\dfrac{18}{5}\) - \(\dfrac{8}{3}\)

   x\(\dfrac{7}{12}\)                   = \(\dfrac{14}{15}\)

  x                         = \(\dfrac{14}{15}\) : \(\dfrac{7}{12}\)

 x                          = \(\dfrac{8}{5}\)

Câu 1:

\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)

\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)

\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)

\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)

\(\Leftrightarrow50x-16=0\)

\(\Leftrightarrow50x=16\)

\(\Leftrightarrow x=\dfrac{8}{25}\)

Câu 2 :

\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)

<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)

<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)

<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x

<=> 11x+27 = 26x -5

<=> ( 26x - 5 ) - ( 11x + 27 ) = 0

<=> 15x - 32 = 0

<=> 15x = 32

<=> x = \(\dfrac{32}{15}\)

Câu 3:

x - 4/3 - 3x - 1/12 = 3x + 1/4 + 9x - 2/8

<=> 4x - 16 - 3x + 1/12 = 6x + 2 + 9x - 2/8

<=> x - 15/12 = 15x/8

<=> 8x - 120 = 180x

<=> 120 = -172x <=> x = -172/120 = -43/30

\(a,\Leftrightarrow\left[{}\begin{matrix}-\dfrac{4}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\-\dfrac{4}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\\ c,\Leftrightarrow\left(\dfrac{1}{2}\right)^x\left(1+\dfrac{1}{4}\right)=\dfrac{5}{4}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^x=1\Leftrightarrow x=0\)

b: Ta có: \(3^x+3^{x+2}=20\)

\(\Leftrightarrow3^x\cdot10=20\)

\(\Leftrightarrow3^x=2\left(loại\right)\)

a)4/5+x=2/3

x=2/3-4/5

x=-2/15

b)-5/6-x=2/3

x=-5/6-2/3

x=-3/2

c)1/2x+3/4=-3/10

1/2x=-3/10-3/4

1/2x=-21/20

x=-21/20:1/2

x=-21/10

d)x/3-1/2=1/5

x/3=1/5+1/2

x/3=7/10

10x/30=21/30

10x=21

x=21:10

x=21/10

a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)

\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)

c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)

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

\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)

b: \(=\dfrac{a+2b}{3a-b}-\dfrac{2a-5b}{3a-b}\)

\(=\dfrac{a+2b-2a+5b}{3a-b}=\dfrac{-a+7b}{3a-b}\)

c: \(=\dfrac{2+x-3}{\left(x+3\right)\left(x-3\right)}=\dfrac{x+1}{\left(x+3\right)\left(x-3\right)}\)

d: \(=\dfrac{4x+x^2-2x+2x+4}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2+4x+4}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{x-2}\)

e: \(=\dfrac{3x^2-x+3+1-2x+x^2-2x^2-2x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\dfrac{2x^2-5x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)

Câu 1: 

=>15(2x+1)-8(3x-1)=100

=>30x+15-24x+8=100

=>6x+23=100

hay x=77/6

Câu 2:

=>2(5x-3)+12-3(7x-1)=x+2

=>10x-6+12-21x+3-x-2=0

=>-12x=-7

hay x=7/12

Câu 3: 

\(\Leftrightarrow2\left(x^2-1\right)+3\left(x+1\right)=2\left(x^2-4x+4\right)\)

\(\Leftrightarrow2x^2-2+3x+3-2x^2+8x-8=0\)

=>11x-7=0

hay x=-7/11

Câu 4:

(x - 4)^3/6 + 1 = x(x + 1)/2 - (x - 5)(x + 5)/3

<=> (x - 4)^3 + 6/6 = x^2 + x/2 - x^2 - 25/3

<=> (x - 4)^3 + 6/6 = 3x^2 + 3x - 2x^2 + 50/6

<=> (x - 4)^3 + 6 = 3x^2 + 3x - 2x^2 + 50

<=> x^3 - 12x^2 + 48x - 58 = x^2 + 3x + 50

<=> x^3 -13x^2 + 45x - 108 = 0

Đến đây bạn bấm máy nhẩm nghiệm là ra nhé

Câu 5:

3(x + 2)^3/5 - (x - 1)^2/10 = (x - 3)(x + 3)/2

<=> 6(x + 2)^3 - (x - 1)^2/10 = 5(x^2 - 9)/10

<=> 6(x + 2)^3 - (x - 1)^2 = 5(x^2 - 9)

<=> 6x^3 + 36x^2 + 72x + 48 - x^2 + 2x - 1 - 5x^2 + 45 = 0

<=> 6x^3 + 30x^2 + 74x + 92 = 0

Đến đây bạn bấm máy nhẩm nghiệm như câu 4 nhé

a,sửa đề : đk x khác -2;  2 

 \(x^2+x-2+5x-10=12+x^2-4\)

\(\Leftrightarrow6x-20=0\Leftrightarrow x=\dfrac{10}{3}\left(tm\right)\)

b, \(3x-12+5+5x=105\Leftrightarrow8x=112\Leftrightarrow x=14\)

c, \(3x^2+14x-49=-\left(x^2+2x-15\right)\)

\(\Leftrightarrow4x^2+16x-34=0\Leftrightarrow x=\dfrac{-4\pm5\sqrt{2}}{2}\)

a. ko hỉu đề lắm :v

b.\(\dfrac{x-4}{5}+\dfrac{1+x}{3}=7\)

\(\Leftrightarrow\dfrac{3\left(x-4\right)+5\left(1+x\right)}{15}=\dfrac{105}{15}\)

\(\Leftrightarrow3\left(x-4\right)+5\left(1+x\right)=105\)

\(\Leftrightarrow3x-12+5+5x-105=0\)

\(\Leftrightarrow8x-112=0\)

\(\Leftrightarrow8x=112\)

\(\Leftrightarrow x=14\)

c.\(\left(3x-7\right)\left(x+7\right)=\left(5+x\right)\left(3-x\right)\)

\(\Leftrightarrow3x^2+21x-7x-49=15-5x+3x-x^2\)

\(\Leftrightarrow4x^2+16x-64=0\)

Nghiệm xấu lắm bạn