a,-4/7=x/21
-12/21 = x/21
x= -12 

b,(x-3)/15=1/-5
x - 3 = -1/5 * 15
x - 3 = -3
x = 0

c,.(3x+8)/-12=-5/30
=> 3x + 8 = 2
=> 3x=-6
=>x=-2

a) (2/5 x 25/29) + (3/5 x 25/29)
= (50/145) + (75/145)
= 125/145

b) (5/2 x 3/7) - (3/14 : 6/7)
= 15/14 - (3/14 x 7/6)
= 15/14 - 1/2
= (30/28) - (14/28)
= 16/28
= 4/7

c) (15/4 : 5/12) - (6/5 : 11/15)
= (15/4 x 12/5) - (6/5 x 15/11)
= 180/20 - 90/55
= 9 - 18/11
= (99/11) - (18/11)
= 81/11
= 7 4/11

a) (2/3) + (20/21 x 3/2 x 7/5)
= 2/3 + (60/210)
= 2/3 + 2/7
= (14/21) + (6/21)
= 20/21

b) (5/17 x 21/32 x 47/24 x 0)
= 0

c) (11/3 x 26/7) - (26/7 x 8/3)
= (286/21) - (208/21)
= 78/21
= 3 9/21
= 3 3/7

a) (25/8) : x = 5/16
=> (25/8) x (16/5) = x
=> 4 = x

b) x + (7/15) = 6/15
=> x = (6/15) - (7/15)
=> x = -1/15

c) x : (28/49) = 7/12
=> x x (49/28) = 7/12
=> x = (7/12) x (28/49)
=> x = 1/2

a) 6 x x = (5/8) : (3/4)
=> 6x = (5/8) x (4/3)
=> 6x = 20/24
=> 6x = 5/6
=> x = (5/6) / 6
=> x = 5/36

câu,b,không,đủ,thông,tin,nhan,bạn.

a: \(\dfrac{16}{15}\cdot\dfrac{-5}{14}\cdot\dfrac{54}{24}\cdot\dfrac{56}{21}\)

\(=\dfrac{16}{24}\cdot\dfrac{-5}{15}\cdot\dfrac{54}{21}\cdot\dfrac{56}{14}\)

\(=\dfrac{2}{3}\cdot\dfrac{-1}{3}\cdot\dfrac{18}{7}\cdot4\)

\(=\dfrac{-2\cdot2\cdot4}{7}=-\dfrac{16}{7}\)

b: \(\dfrac{7}{3}\cdot\dfrac{-5}{2}\cdot\dfrac{15}{21}\cdot\dfrac{4}{-5}\)

\(=\dfrac{7}{3}\cdot\dfrac{5}{7}\cdot\dfrac{5}{2}\cdot\dfrac{4}{5}\)

\(=\dfrac{5}{3}\cdot\dfrac{4}{2}=\dfrac{5}{3}\cdot2=\dfrac{10}{3}\)

c: \(\left(-\dfrac{2}{5}+\dfrac{1}{3}\right)\left(\dfrac{3}{2}-\dfrac{3}{7}\right)\)

\(=\dfrac{-2\cdot5+3}{15}\cdot\dfrac{3\cdot7-3\cdot2}{14}\)

\(=\dfrac{-7}{15}\cdot\dfrac{15}{14}=\dfrac{-7}{14}=-\dfrac{1}{2}\)

d: \(\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\left(5-\dfrac{1}{4}\right)\)

\(=\dfrac{3-2}{6}\cdot\dfrac{20-1}{4}\)

\(=\dfrac{1}{6}\cdot\dfrac{19}{4}=\dfrac{19}{24}\)

e: \(\left(\dfrac{4}{5}\right)^2=\dfrac{4^2}{5^2}=\dfrac{16}{25}\)

a: 2x-261=-21

=>\(2x=-21+261=240\)

=>\(x=\dfrac{240}{2}=120\)

b: \(15\cdot\left(-2\right)-x=45\)

=>-30-x=45

=>x=-30-45=-75

c: \(-18+\left(7-x\right)^2=-2\)

=>\(\left(x-7\right)^2-18=-2\)

=>\(\left(x-7\right)^2=-2+18=16\)

=>\(\left[{}\begin{matrix}x-7=4\\x-7=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=3\end{matrix}\right.\)

d: \(2x-1⋮x+1\)

=>\(2x+2-3⋮x+1\)

=>\(-3⋮x+1\)

=>\(x+1\inƯ\left(-3\right)\)

=>\(x+1\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{0;-2;2;-4\right\}\)

biết j mới dk bn??

(x^2-1).(x^2-3).(x^2-5).(x^2-7)\(\le\)0

Biết j vậy phải có điều kiện chứ ...

|x-2|=3x       

\(\Rightarrow\orbr{\begin{cases}x-2=3x\\x-2=-3x\end{cases}}\)  

\(\Rightarrow\orbr{\begin{cases}-2=3x-x\Rightarrow-2x=2\Rightarrow x=-1\\x-2=-3x\Rightarrow x+3x=2\Rightarrow4x=2\Rightarrow x=2\end{cases}}\)

Vậy x=-1 hoặc x=2

|x-3|-7=13

|x-3|=13+7

|x-3|=20

\(\Rightarrow\orbr{\begin{cases}x-3=20\\x-3=-20\end{cases}}\)        

\(\Rightarrow\orbr{\begin{cases}x=20+3=23\\x=-20+3=-17\end{cases}}\)

Vậy x=23 hoặc x=-17

72-(43-|x|)=45

      43-|x|=72-45

      43-|x|=27 

           |x|=43-27

           |x|=16

=> x=+16

Vậy x=+16

3|x-1|-5=7 

3|x-1|=7+5

3|x-1|=12

  |x-1|=12:3

  |x-1|=4

\(\Rightarrow\orbr{\begin{cases}x-1=4\Rightarrow x=4+1=5\\x-1=-4\Rightarrow x=-4+1=-3\end{cases}}\)     

Vậy x=5 hoặc x=-3

(7-x )(x+9)=0     

* 7-x=0             * x+9=0

     x=7-0                 x=0-9

     x=7                    x=-9

Vậy x=7 hoặc x=-9

(x-2)(x+15)=0

* x-2=0               * x+15=0

      x=0+2                   x=0-15

      x=2                       x=-15

Vậy x=2 hoặc x=-15

-12(x-5)+7(3-x)=5   

-12x+60+21-7x=5

-12x-7x=5-60-21

-19x=-76

     x=-76:(-19)    

     x=4

Vậy x=4

2x²-3=29

2x²=29+3

2x²=32

  x²=32:2

  x²=16

  x²=4²

=> x=4

Vậy x=4

\(\left|x-3\right|-7=13\)

\(\left|x-3\right|=20\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=20\\x-3=-20\end{cases}\Leftrightarrow\orbr{\begin{cases}x=23\\x=-17\end{cases}}}\)

vậy x=23 hoặc x=-17

\(\left(7-x\right)\left(x+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}7-x=0\\x+9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-9\end{cases}}}\)

vậy x=7 hoặc x=-9

\(\left(x-2\right)\left(x+15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-15\end{cases}}}\)

vậy x=2 hoặc x=-15

\(2x^2-3=29\)

\(2x^2=32\)

\(x^2=16\)

\(\Rightarrow x=\pm4\)

a) x : 1/4 = 7/5
    x         = 7/5 x 1/4
    x         = 7/20
b) 9/2 - x = 3/5
            x = 9/2 - 3/5
            x = 39/10
c) 12/5 : x = 7/2
              x = 12/5 : 7/2
              x = 24/35

a) \(x\div\dfrac{1}{4}=\dfrac{7}{5}\)

    \(x=\dfrac{7}{5}\times\dfrac{1}{4}\)

    \(x=\dfrac{7}{20}\)

b) \(\dfrac{9}{2}-x=\dfrac{3}{5}\)

            \(x=\dfrac{9}{2}-\dfrac{3}{5}\)

            \(x=\dfrac{39}{10}\)

c) \(\dfrac{12}{5}\div x=\dfrac{7}{2}\)

             \(x=\dfrac{12}{5}\div\dfrac{7}{2}\)

             \(x=\dfrac{12}{5}\times\dfrac{2}{7}\)

             \(x=\dfrac{24}{35}\)

a)\(x=\dfrac{1}{4}\)x\(\dfrac{7}{5}=\dfrac{7}{20}\)

b)\(x=\dfrac{9}{2}-\dfrac{3}{5}=\dfrac{39}{10}\)

c)\(x=\dfrac{12}{5}:\dfrac{7}{2}=\dfrac{24}{35}\)

2) \(\left(x-2\right).\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)

vậy \(x=2\) hoặc    \(x=-4\)

3) \(\left(x-2\right).\left(x+15\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+15=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-15\end{cases}}}\)

vậy \(x=2\) hoặc   \(x=-15\)

4) \(\left(7-x\right).\left(x+19\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7-x=0\\x+19=0\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=-19\end{cases}}}\)

vậy \(x=7\) hoặc    \(x=-19\)

8) \(2x^2-3=29\)

\(2x^2=29+3\)

\(2x^2=32\)

\(x^2=32\div2\)

\(x^2=16\)

\(\Rightarrow\orbr{\begin{cases}x^2=4^2\\x^2=\left(-4\right)^2\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

vậy \(x=4\) hoặc    \(x=-4\)

7)Ta có : x - 3 = 0                                                 x - 5 = 0

           => x     = 0 + 3                                        => x    = 0 +5 

           => x     =    3                                            => x   = 5

Ta lập bảng xét dấu : 

Vậy để (x-3).(x-5) < 0 => 3<x<5 => x = 4

6) | x | <3

=>x thuộc cộng trừ 1 , cộng trừ 2