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

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

Ta có : x² - 2x - (x + 3)² = 6

<=> x² - 2x - x² - 6x - 9 = 6

<=> -8x - 9 = 6 

=> -8x = 15

=> x = \(\frac{15}{-8}\)

a, ( 8x - 3 ) ( 3x + 2 ) - ( 4x + 7 ) ( x + 4 ) = ( 2x + 1 ) ( 5x - 1 )

 ( 24x² + 16x - 9x - 6 ) - ( 4x² - 16x - 7x + 28 ) = 10x² - 2x + 5x -1

24x² + 16x - 9x - 6 -4x² - 16x - 7x - 10x² + 2x - 5x = 6 + 28 - 1

10x² -19x = 33

10x² - 19x -33 = 0 \(\Leftrightarrow\)10x( x+ 3 ) + 11 ( x- 3 ) = 0

=>  ( x- 3 ) ( 10x + 11 ) = 0\(\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-11}{10}\end{cases}}\)

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

4( x² - 5x - x + 5 ) - ( x² + 5x + 2x + 10 ) = 3( x² + 2x - x - 2 )

4x² - 20x - 4x + 20 - x² - 5x - 2x - 10 = 3x² + 6x - 3x - 6

( 4x² - x² ) + ( -20x - 4x - 5x - 2x ) + 20 - 10 = 3x² + ( 6x - 3x ) - 6

3x² - 31x - 3x² - 3x = -6-10

-34x = -16

x = \(\frac{8}{17}\)

a) \(\frac{x-5}{7}=\frac{5}{3}\)                                                                         

\(\Rightarrow\left(x-5\right)\cdot3=7\cdot5\)

\(\Rightarrow3x-15=35\)

\(\Rightarrow3x=15+35\)

\(\Rightarrow3x=50\)

\(\Rightarrow x=\frac{50}{3}\)

b) \(\frac{x-3}{3}=\frac{12}{x-3}\)

\(\Rightarrow\left(x-3\right)\cdot\left(x-3\right)=3\cdot12\)

\(\Rightarrow\left(x-3\right)^2=36\)

\(\Rightarrow\left(x-3\right)^2=6^2\)hoặc \(\left(x-3\right)^2=\left(-6\right)^2\)

\(\Rightarrow x-3=6\)                      \(x-3=-6\)

\(\Rightarrow x=6+3\)                       \(x=-6+3\)

\(\Rightarrow x=9\)                    hoặc     \(x=-3\)

c) \(\frac{5-2x}{4}=\frac{7}{3}\)

\(\Rightarrow\left(5-2x\right)\cdot3=4\cdot7\)

\(\Rightarrow15-6x=28\)

\(\Rightarrow6x=15-28\)

\(\Rightarrow6x=-13\)

\(\Rightarrow x=-\frac{13}{6}\)

d) \(\frac{x+1}{4}=\frac{7}{3}\)

\(\Rightarrow\left(x+1\right)\cdot3=2\cdot7\)

\(\Rightarrow3x+3=28\)

\(\Rightarrow3x=28-3\)

\(\Rightarrow3x=25\)

\(\Rightarrow x=\frac{25}{3}\)

Chúc bạn học tốt !!!

a) -3/5

b) -9/4

c) x thuộc N*( chắc thế)

Bn giải kĩ đc k 

3(x + 2)^2 + (2x - 1)^2 - 7(x + 3)(x - 3) = 36

=> 3(x^2 + 4x + 4) + 4x^2 - 4x + 1 - 7(x^2 - 9) = 36

=> 3x^2 + 12x + 12 + 4x^2 - 4x + 1 - 7x^2 + 63 = 36

=> 8x + 76 = 36

=> 8x = -40

=> x = -5

\(x+\dfrac{1}{x}=3\Leftrightarrow\left(x+\dfrac{1}{x}\right)^3=27\\ \Leftrightarrow x^3+\left(\dfrac{1}{x}\right)^3+3x\cdot\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)=27\\ \Leftrightarrow x^3+\dfrac{1}{x^3}+3\cdot3=27\\ \Leftrightarrow x^3+\dfrac{1}{x^3}=18\)