a, 2\(x\) + 4 - 5\(x\) = -11

    -(5\(x\) - 2\(x\)) + 4 = -11

    -3\(x\)         + 4 = -11

     3\(x\)              = 11 + 4

     3\(x\)              = 15

       \(x\)              = 15 : 3

        \(x\)              = 5

b, \(x\) - (-5) = 8

    \(x\) + 5 = 8

    \(x\)       = 8 - 5

    \(x\)       = 3

c, (\(x\) + 2)² = 25

   \(\left[{}\begin{matrix}x+2=5\\x+2=-5\end{matrix}\right.\)

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

B2

( a³ + a²b + ab² + b³ ).( a - b ) = a⁴ - b⁴

[( a³ + b³ + ab.( a + b )].( a - b ) = a⁴ - b⁴

[( a + b ).( a² - ab + b² ) + ab.( a + b )].( a - b ) = a⁴ - b⁴

 ( a + b ).( a² - ab + b² + ab ).( a - b ) = a⁴ - b⁴

( a + b ).( a² + b² ).( a  -  b ) = a⁴ - b⁴

 ( a² - b² ).( a² + b² ) = a⁴ - b⁴

 a⁴ - b⁴ = a⁴ - b⁴  ( đpcm )

a) \(-\frac{3}{4}x+\frac{1}{6}x=1-2\frac{5}{9}\)
    \(\left(-\frac{3}{4}+\frac{1}{6}\right).x=1-\frac{23}{9}\)
                     \(-\frac{7}{12}.x=-\frac{14}{9}\)
                                  \(x=-\frac{14}{9}:\left(-\frac{7}{12}\right)\)
                                  \(x=\frac{8}{3}\)
Vậy x = ...
b) \(\left|2x-\frac{3}{8}\right|+2\frac{3}{4}=3\frac{1}{16}\)
     \(\left|2x-\frac{3}{8}\right|+\frac{11}{4}=\frac{49}{16}\)
                    \(\left|2x-\frac{3}{8}\right|=\frac{49}{16}-\frac{11}{4}\)      
                    \(\left|2x-\frac{3}{8}\right|=\frac{5}{16}\)
 \(\Rightarrow\left|2x-\frac{3}{8}\right|\in\text{{}\frac{5}{16};-\frac{5}{16}\)}
Nếu, \(2x-\frac{3}{8}=\frac{5}{16}\)
                     \(2x=\frac{11}{16}\)
                        \(x=\frac{11}{32}\)
Nếu, \(2x-\frac{3}{8}=-\frac{5}{16}\)
                     \(2x=\frac{1}{16}\)
                        \(x=\frac{1}{32}\)
Vậy \(x\in\text{{}\frac{1}{32};\frac{11}{32}\)}
                     

a) 5 - 4x = 3x - 9

\(\Leftrightarrow5-4x-3x+9=0\)

\(\Leftrightarrow14-7x=0\)

\(\Leftrightarrow7x=14\Leftrightarrow x=2\)

Vậy \(S=\left\{2\right\}\)

b) \(\left(x-4\right)\left(3x+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

Vậy \(S=\left\{-3;4\right\}\)

c) \(\dfrac{x}{x+4}+\dfrac{12}{x-4}=\dfrac{4x+48}{x\cdot x-16}\)(1)

ĐKXĐ: \(x\ne\pm4\)

\(\left(1\right)\Leftrightarrow\dfrac{x\left(x-4\right)+12\left(x+4\right)-4x-48}{\left(x+4\right)\left(x-4\right)}=0\)

\(\Leftrightarrow x^2-4x+12x+48-4x-48=0\)

\(\Leftrightarrow x^2+4x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-4\left(KTM\right)\end{matrix}\right.\)

Vậy \(S=\left\{0\right\}\)

d) \(4-2x=7-x\)

\(\Leftrightarrow4-2x-7+x=0\)

\(\Leftrightarrow-x-3=0\)

\(\Leftrightarrow-x=3\Leftrightarrow x=-3\)

Vậy \(S=\left\{-3\right\}\)

e) \(\left(x+4\right) \left(8-4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\8-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{-4;2\right\}\)

f) \(\dfrac{x}{x+5}+\dfrac{11}{x-5}=\dfrac{x+55}{x\cdot x-25}\left(2\right)\)

ĐKXĐ: \(x\ne\pm5\)

\(\left(2\right)\Leftrightarrow\dfrac{x\left(x-5\right)+11\left(x+5\right)-x-55}{\left(x+5\right)\left(x-5\right)}=0\)

\(\Leftrightarrow x^2-5x+11x+55-x-55=0\)

\(\Leftrightarrow x^2+5x=0\)

\(\Leftrightarrow x\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-5\left(KTM\right)\end{matrix}\right.\)

Vậy \(S=\left\{0\right\}\)

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

\(\Leftrightarrow\dfrac{3\left(3x+2\right)-3x-1-10-12x}{6}=0\)

\(\Leftrightarrow9x+6-3x-1-10-12x=0\)

\(\Leftrightarrow-6x-5=0\)

\(\Leftrightarrow-6x=5\)

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

Vậy \(S=\left\{-\dfrac{5}{6}\right\}\)

h) \(2x-\left(3-5x\right)=4\left(x+3\right)\)

\(\Leftrightarrow2x-3+5x-4x-12=0\)

\(\Leftrightarrow3x-15=0\)

\(\Leftrightarrow x=5\)

Vậy \(S=\left\{5\right\}\)

i) \(3x-6+x=9-x\)

\(\Leftrightarrow3x-6+x-9+x=0\)

\(\Leftrightarrow5x-15=0\)

\(\Leftrightarrow x=3\)

Vậy \(S=\left\{3\right\}\)

k)\(2t-3+5t=4t+12\)

\(\Leftrightarrow2t-3+5t-4t-12=0\)

\(\Leftrightarrow3t-15=0\)

\(\Leftrightarrow t=5\)

Vậy \(S=\left\{5\right\}\)

18 - 4\(x\) = -20 - 6\(x\) 

-4\(x\) + 6\(x\) = - 20  - 18

      2\(x\)    = - 38

         \(x\)  = - 19

h, -15 \(\times\) 24 = -7\(x\) + 32

     7\(x\) = 360 + 32

      7\(x\) = 392

         \(x\) =  392:7

          \(x\) = 56

i, 15\(x\) -3.(4\(x\) - 6) = -12 + 36

    15\(x\) - 12\(x\) + 18 =  24

      3\(x\) = 24 - 18

       3\(x\) = 6

         \(x\) = 2

k, -10\(x\) - 27 = -7\(x\) + 33

    -27 - 33 = -7\(x\) + 10\(x\)

     3\(x\) = -60

        \(x\) = -20

m, -17\(x\) - 24 = - 9\(x\) - 40

     - 24 + 40 =  -9\(x\) + 17\(x\)

          8\(x\) = 16

             \(x\) = 2

n, -23 \(\times\) 25 = -18\(x\) + 75

     -575 = -18\(x\) + 75

       18\(x\) = 575 + 75

        18\(x\) = 650

              \(x\) = \(\dfrac{325}{9}\)

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 \(x=\dfrac{5}{6}\)

câu a không có về phải=> k làm dc

b)x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12

<=> 3x^2 +2x +x^2+2x+1 - 4x^2 +25 +12=0

<=> 4x+38=0

=>4x= -38

=>x= -38/4= -19/2

d)x^2-x=0

<=>x(x-1)=0

=>x=0 hoặc x-1=0

=>x=0 hoặc x=1