a) ( x - 1 )( x² + x + 1 ) + x( x + 2 )( 2 - x ) = 5

<=> x³ - 1 - x( x + 2 )( x - 2 ) = 5 

<=> x³ - 1 - x( x² - 4 ) = 5

<=> x³ - 1 - x³ + 4x = 5

<=> 4x - 1 = 5

<=> 4x = 6

<=> x = 6/4 = 3/2

b) 5x( x - 3 )² - 5( x - 1 )³ + 15( x + 4 )( x - 4 ) = 5

<=> 5x( x² - 6x + 9 ) - 5( x³ - 3x² + 3x - 1 ) + 15( x² - 16 ) = 5

<=> 5x³ - 30x² + 45x - 5x³ + 15x² - 15x + 5 + 15x² - 240 = 5

<=> 30x - 235 = 5

<=> 30x = 240

<=> x = 8

a,\(\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)

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

\(< =>x^3-1+4x-x^3=5\)

\(< =>4x-1-5=0< =>4x-6=0< =>x=\frac{3}{2}\)

b, \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+4\right)\left(x-4\right)=5\)

\(< =>5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-16\right)=5\)

\(< =>5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-240=5\)

\(< =>\left(5x^3-5x^3\right)+\left(15x^2+15x^2-30x^2\right)+\left(45x-15x\right)+5-240=5\)

\(< =>30x-240=5-5=0< =>x=\frac{24}{3}=8\)

a, \(\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)

\(\Leftrightarrow x^3-1+4x-x^3=5\Leftrightarrow-6+4x=0\Leftrightarrow x=\frac{3}{2}\)

b, \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+4\right)\left(x-4\right)=5\)

\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-240=5\)

\(\Leftrightarrow30x-240=5\Leftrightarrow x=8\)

\(\left(x-\dfrac{3}{2}\right)\times\left(2x+1\right)>0\)

Th1:

\(x-\dfrac{3}{2}>0\Leftrightarrow x>\dfrac{3}{2}\)

\(2x+1>0\Leftrightarrow2x>1\Leftrightarrow x>\dfrac{1}{2}\)

( 1 )

Th2: 

\(x-\dfrac{3}{2}< 0\Leftrightarrow x< \dfrac{3}{2}\)

\(2x+1< 0\Leftrightarrow2x< -1\Leftrightarrow x< -\dfrac{1}{2}\)

( 2 )

Từ ( 1 ) và ( 2 ), ta có:

\(\Rightarrow x< -\dfrac{1}{2};x>\dfrac{3}{2}\)

\(\left(2-x\right)\times\left(\dfrac{4}{5}-x\right)< 0\)

Th1:

\(2-x>0\Leftrightarrow x>2\)

\(\dfrac{4}{5}-x< 0\Leftrightarrow x< \dfrac{4}{5}\)

( Loại )

Th2:

\(2-x< 0\Leftrightarrow x< 2\)

\(\dfrac{4}{5}-x>0\Leftrightarrow x>\dfrac{4}{5}\)

=> \(\dfrac{4}{5}< x< 2\)

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

=> 3 ( x + 2 ) = 4 ( x - 5 ) 

=> 3x + 6 = 4x - 20

=> 3x - 4x = - 6 - 20

=> - 1x = - 26

=> x = 26

1) 3/x-5=-4/x+2

     3x+6=-4x+20

     7x .  =14

      x  .   =2

2) x+3/-4=-9/x+3

   (x+3)^2=36

  Ta có hai trường hợp:

*x+3=6=)x=3

*x+3=-6=)x=-9

3) (34/5-2x). 11/3=55/7

     34/5-2x .          =15/7

      75-30x.            =238

           30x.            = -163

               x.            =-163/30

a: \(\Leftrightarrow x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)

mà -3<x<30

nên \(x\in\left\{-2;-1;1;2;3;4;6;9;12;18\right\}\)

b: \(\Leftrightarrow x\in\left\{0;4;-4;8;-8;12;-12;...\right\}\)

mà -16<=x<20

nên \(x\in\left\{-16;-12;-8;-4;0;4;8;12;16\right\}\)

c: \(\Leftrightarrow x-1+4⋮x-1\)

\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)

d: \(\Leftrightarrow2x+4-5⋮x+2\)

\(\Leftrightarrow x+2\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{-1;-3;3;-7\right\}\)

Bài 1:

a) \(x.\dfrac{3}{4}=\dfrac{9}{14}\)

\(\Rightarrow x=\dfrac{9}{14}:\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{6}{7}\)

b) \(x:\dfrac{5}{9}=\dfrac{3}{10}\)

\(\Rightarrow x=\dfrac{3}{10}.\dfrac{5}{9}\)

\(\Rightarrow x=\dfrac{1}{6}\)

help me

Bài 2:

a) \(\dfrac{3}{10}+\dfrac{2}{5}:\dfrac{4}{15}=\dfrac{3}{10}+\dfrac{3}{2}=\dfrac{9}{5}\)

b) \(\dfrac{4}{7}.3+\dfrac{2}{3}.2=\dfrac{12}{7}+\dfrac{4}{3}=\dfrac{64}{21}\)

\(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|=0\)

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

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

Vậy \(S=\varnothing\)

b: \(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|\ge0\forall x\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

1) \(\left(x-3\right)^2-4=0\)

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

\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

2) \(x^2-2x=24\)

\(\Leftrightarrow x^2-2x-24=0\)

\(\Leftrightarrow x^2+4x-6x-24=0\)

\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Câu 3 số xấu rồi e

alo =0 nhé anh pop pop

a: x/2=-5/y

=>xy=-10

=>\(\left(x,y\right)\in\left\{\left(1;-10\right);\left(-10;1\right);\left(-1;10\right);\left(10;-1\right);\left(2;-5\right);\left(-5;2\right);\left(-2;5\right);\left(5;-2\right)\right\}\)

b: =>xy=12

mà x>y>0

nên \(\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)

c: =>(x-1)(y+1)=3

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

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

d: =>y(x+2)=5

=>\(\left(x+2;y\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-1;5\right);\left(3;1\right);\left(-3;-5\right);\left(-7;-1\right)\right\}\)