\(|\dfrac{4}{3}x-\dfrac{3}{4}|=\left|-\dfrac{1}{3}\right|.\left|x\right|\Leftrightarrow|\dfrac{4}{3}x-\dfrac{3}{4}|=\dfrac{1}{3}.\left|x\right|\left(1\right)\)

Tìm nghiệm \(\dfrac{4}{3}x-\dfrac{3}{4}=0\Leftrightarrow\dfrac{4}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.\dfrac{3}{4}\Leftrightarrow x=\dfrac{9}{16}\)

                    \(x=0\)

Lập bảng xét dấu :

     \(x\)                           \(0\)                   \(\dfrac{9}{16}\)

\(\left|\dfrac{4}{3}x-\dfrac{3}{4}\right|\)         \(-\)       \(0\)           \(-\)       \(0\)        \(+\)

      \(\left|x\right|\)              \(-\)       \(0\)           \(+\)       \(0\)        \(+\)

TH1 : \(x< 0\)

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

\(\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=-\dfrac{1}{3}.x\)

\(\Leftrightarrow\dfrac{4}{3}x-\dfrac{1}{3}x=\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{3}{4}\) (loại vì không thỏa \(x< 0\))

TH2 : \(0\le x\le\dfrac{9}{16}\)

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

\(\Leftrightarrow\dfrac{4}{3}x+\dfrac{1}{3}x=\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.\dfrac{3}{5}\Leftrightarrow x=\dfrac{9}{20}\) (thỏa điều kiện \(0\le x\le\dfrac{9}{16}\))

TH3 : \(x>\dfrac{9}{16}\)

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

\(\Leftrightarrow\dfrac{4}{3}x-\dfrac{1}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}\) (thỏa điều kiện \(x>\dfrac{9}{16}\))

Vậy \(x\in\left\{\dfrac{9}{20};\dfrac{3}{4}\right\}\)

5)

để \(\frac{5x-3}{x+1}\)là số nguyên

\(5x-3⋮x+1\)

\(x+1⋮x+1\)

\(\Rightarrow5\left(x+1\right)⋮x+1\)

\(5x-3-\left(5x-5\right)⋮x+1\)

\(-2⋮x+1\)

\(\Rightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Vậy \(x\in\left\{0;-2;1;-3\right\}\)

a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0

=>-5x-4=0

=>x=-4/5

b: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81

=>83x=83

=>x=1

a: ĐKXĐ: \(x>0\)

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

\(=x+\sqrt{x}-2\sqrt{x}-1+1\)

\(=x-\sqrt{x}\)

A.  2.\(|3x+1|\)=\(\frac{3}{4}\)-\(\frac{5}{8}\)

     2.\(|3x+1|\)=1/8

        \(|3x+1|\)=1/8:2

        \(|3x+1|\)=1/16

TH1 : 3x+1=1/16

         3x=1/16-1

         3x=-15/16

         x=-15/16:3

          x=-5/16

a,\(\frac{3}{4}-2.\left|3x+1\right|=\frac{5}{8}\)

\(\Rightarrow2.\left|3x+1\right|=\frac{3}{4}-\frac{5}{8}=\frac{6}{8}-\frac{5}{8}=\frac{1}{8}\)

\(\Rightarrow\left|3x+1\right|=\frac{1}{8}.\frac{1}{2}=\frac{1}{16}\)

\(\Rightarrow\orbr{\begin{cases}3x+1=\frac{1}{16}\\3x+1=\frac{-1}{16}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3x=\frac{1}{16}-1=\frac{-15}{16}\\3x=\frac{-1}{16}-1=\frac{-17}{16}\end{cases}}\)

                                          \(\Rightarrow\orbr{\begin{cases}x=\frac{-15}{16}.\frac{1}{3}=\frac{-5}{16}\\x=\frac{-17}{16}.\frac{1}{3}=\frac{-17}{48}\end{cases}}\)

Vậy....

b,\(\left|3x+2\right|-\left|x-3\right|=\frac{7}{2}\left(1\right)\)

Ta có bảng xét dấu

Nếu x<\(\frac{-2}{3}\)       thì \(\left|3x+2\right|-\left|x-3\right|\) \(=-3x-2-3+x\)

                                                                         \(=-2x-5\)

Từ (1) \(\Rightarrow-2x-5=\frac{7}{2}\)

          \(\Rightarrow-2x=\frac{7}{2}+5=\frac{17}{2}\)

           \(\Rightarrow x=\frac{17}{2}\cdot\frac{-1}{2}=\frac{-17}{4}\)(thỏa mãn x<\(\frac{-2}{3}\)

Nếu \(\frac{-2}{3}\le x\le3\)thì \(\left|3x+2\right|-\left|x-3\right|=3x+2-\left(3-x\right)\)

                                                                                \(=3x+2-3+x\)

                                                                                 \(=2x-1\)

Từ (1)\(\Rightarrow\)\(2x-1=\frac{7}{2}\)

    \(\Rightarrow2x=\frac{9}{2}\)

      \(\Rightarrow x=\frac{9}{4}\)(thỏa mãn......

Còn trưonwfg hợp cuối bạn tự làm nốt nhé

4) (3x-2)(x-3)= 3x(x-3)-2(x-3)

=3x.x+3x.(-3)-2.x-2.(-3)

=\(3x^2\)-9x-4x+6

=\(3x^2\)+(-9x-4x)+6

=\(3x^2\)-13x+6

5) (2x+1)(x+3)=2x(x+3)+1(x+3)

=2x.x+2x.3+1.x+1.3

=\(2x^2\)+6x+1x+3

=\(2x^2\)+(6x+1x)+3

=\(2x^2\)+7x+3

6) (x-3)(3x-1)=x(3x-1)-3(3x-1)

=x.3x+x.(-1)-3.3x-3.(-1)

=\(3x^2\)-1x-9x+3

=\(3x^2\)+(-1x-9x)+3

=\(3x^2\)-10x+3

rút gọn biểu thức

A) \(x^2\)-(x+4)(x-1)=\(x^2\)- x(x-1)-4(x-1)

=\(x^2\)-x.x-x.(-1)-4.x-4.(-1)

=\(x^2\)-\(x^2\)+1x-4x+4

=(\(x^2-x^2\))+(1x-4x)+4

= -3x+4

B) x(x+2)-(x-2)(x+4)=x.x+x.2-x(x+4)+2(x+4)

=\(x^2+2x\)-x.x-x.4+2.x+2.4

=\(x^2+2x-x^2-4x+2x+8\)

=(\(x^2-x^2\))+(2x-4x+2x)+8

=8

tính giá trị biểu thức

A=3(x-2)-(2+x)(x-3)

=3.x+3.(-2)-2(x-3)-x(x-3)

=3x-6-2.x-2.(-3)-x.x-x(-3)

=3x-6-2x+6-\(x^2\)+3x

=(3x-2x+3x)+(-6+6)\(-x^2\)

=4x - \(x^2\)

thay x=-8 vào biểu thức thu gọn ta được:

4.(-8)- (-8)\(^2\)

= - 32 +64

= 32

B= x(3-x)-(1+x)(1-x)

=x.3+x.(-x)-1(1-x)-x(1-x)

=3x -\(x^2\)-1.1-1 .(-x)-x.1-x.(-x)

=3x\(-x^2\)-\(1^2\)+1x-1x+\(x^2\)

=(3x+1x-1x)+(\(-x^2+x^2\))-1

=3x-1

thay x=-5 vào biểu thức thu gọn ta được:

3.(-5)-1

=-15-1

=-16

Thu gọn biểu thức

4) (3x - 2) (x - 3) 

= ( 3x² - 2x ) - ( 3x x 3 - 2 x 3 )

= 3x² - 2x - 3x x 3 + 2 x 3

= 3x² - 2x - 9x + 6

= 3x² - 11x + 6 

5) (2x + 1) (x + 3) 

= ( 2x² + 1x ) + ( 6x + 3 )

= 2x² + 1x + 6x + 3

= 2x² + 7x + 3

6) (x - 3) (3x - 1) 

= ( 3x² - 9x ) - ( x - 3 )

= 3x² - 9x - x + 3

= 3x² - 10 + 3

Rút gọn biểu thức

A) x^2 - (x + 4) (x - 1)

= x² - ( x² + 4x ) - ( x + 4 )

= x² - x² - 4x - x - 4

= -5x - 4

B) x (x + 2) - (x - 2) (x + 4)

= x² + 2x - ( x² - 2x ) + ( 4x - 8 )

= x² + 2x - x² + 2x + 4x - 8

= 8x - 8

Tính giá trị biểu thức

A = 3 (x - 2) - (2 + x) (x - 3) tại x = - 8

Thế x = -8 vào, ta có :

= 3 ( -8 -2 ) - ( 2 + -8 ) ( -8 - 3 )

= 3 x ( -10 ) - ( - 6 ) ( -11 )

= -30 - 66

= -96

B = x (3 - x) - (1 + x) ( 1 - x) tại x = - 5

Thế x = - 5 vào, ta có :

= -5 ( 3 - -5 ) - ( 1+ -5 ) ( 1 - -5 )

= -5 x 8 - (-4) x 6

= - 40 - -24

= -40 + 24

= -16

100% đúng 

hok tốt nha 

a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)

    (\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)

     - \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)

     \(x\)   = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))

     \(x=\) - \(\dfrac{1}{2}\)

Vậy \(x=-\dfrac{1}{2}\) 

b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)

           \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)

          \(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)

         3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)

         3\(x\)   - 3,7 = - \(\dfrac{19}{2}\)

         3\(x\)         = - \(\dfrac{19}{2}\) + 3,7

          3\(x\)        = - \(\dfrac{29}{5}\)

           \(x\)         = - \(\dfrac{29}{5}\) : 3

           \(x\)        = - \(\dfrac{29}{15}\)

Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\) 

c; \(\dfrac{7}{9}\) : (2 + \(\dfrac{3}{4}\)\(x\)) + \(\dfrac{5}{9}\) = \(\dfrac{23}{27}\)

    \(\dfrac{7}{9}\): (2 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{23}{27}\) - \(\dfrac{5}{9}\)

     \(\dfrac{7}{9}\):(2 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{8}{27}\)

      2 + \(\dfrac{3}{4}\)\(x\) = \(\dfrac{7}{9}\) : \(\dfrac{8}{27}\)

      2 + \(\dfrac{3}{4}\)\(x\) =  \(\dfrac{21}{8}\)

             \(\dfrac{3}{4}x\) = \(\dfrac{21}{8}\) - 2

             \(\dfrac{3}{4}\)\(x\) =  \(\dfrac{5}{8}\)

               \(x\) = \(\dfrac{5}{8}\) : \(\dfrac{3}{4}\)

              \(x\) =  \(\dfrac{5}{6}\)

Vậy \(x=\dfrac{5}{6}\)