Mình trình bày trong hình ^^ Bn tham khảo nhé

d: Ta có: \(9x^2+6x-8=0\)

\(\Leftrightarrow9x^2+12x-6x-8=0\)

\(\Leftrightarrow\left(3x+4\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

e: Ta có: \(x\left(x-2\right)+x-2=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

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

f: Ta có: \(5x\left(x-3\right)-x+3=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)

\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)

hay \(x=\dfrac{25}{372}\)

Vậy: \(x=\dfrac{25}{372}\)

e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)

f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)

\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)

\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)

\(\Leftrightarrow3x=\dfrac{1}{9}\)

hay \(x=\dfrac{1}{27}\)

g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)

\(\Leftrightarrow\dfrac{4}{3}x=2\)

hay \(x=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)

  Vậy ...

i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)

  Vậy ...

=>12x-33=3^5

=>12x=276

=>x=23

12x-33=9.27

12x-33=243

12x=243+33

12x=276

x=276:12

x=23

12 x- 33= 3^2 . 3^3

12 x- 33= 3^5

12 x- 33=243

12 x     =243+ 33

12 x     = 276

    x     =276:12

    x     = 23

Vậy x = 23

       14,73 : x + 21,27 : x - 12 : x = 33-3x3x3

<=> ( 14,73 + 21,27 - 12 ) : x = 33 - 27 = 6

<=> 24 : x = 6

=> x = 24 : 6 = 4

14,73 : x + 21,27 : x - 12 : x = 33 - 3 * 3 * 3

14,73 : x + 21,27 : x - 12 : x = 6

HAY: ( 14,73 + 21,27 - 12 ) : x = 6

                     24                   : x = 6

                                              x = 24 : 6 

                                              x = 4

chúc bạn học tốt nhé !!! ^_^

vì x xuất hiện nhiều lần nên chúng ta sẽ lấy x làm thừa số chung

[14,73+21,27-12]:x =33-27

     24 : x             =6

            x           =24: 6

            x=4

nhớ nha

\(a,5.\left(x+35\right)=515\)

\(x+35=103\)

\(x=103-35\)

\(x=68\)

\(b,12.x-33=3^2.3^3\)

\(12x-33=243\)

\(12x=276\)

\(x=23\)

\(d,12.\left(x-1\right):3=4^3-2^3\)

\(12.\left(x-1\right):3=56\)

\(12.\left(x-1\right)=168\)

\(x-1=14\)

\(x=15\)

(x-35)-120=0

x-35=120

x=155

156-(x+61)=82

      -(x+61)=-74

        x+61=74

        x=13

12x-33=3^2.3^3

12x-33=243

12x=276

x=23

a: 7x+58=100

nên 7x=42

hay x=6

c: x-56:x=16

nên x-14=16

hay x=30

c)x - 56 : 4  = 16

x - 56       = 16 : 4 

x- 56        = 4

x               =4 + 56

x               = 60 

d)101 + (36 - 4x) = 105 

          (36- 4x ) = 105 - 101 

          36 - 4x = 4

                4x = 36 - 4 

                4x = 32

                  x = 32:4

                  x = 8

a) 58 + 7x = 100

            7x = 100 - 58

            7x = 42

              x = 42 : 7

              x = 6

b) (x - 12) : 12 = 12

              x - 12 = 12 . 12

             x - 12 = 144

                    x = 144 + 12

                    x = 156

e: =>-40+3+33+40-x=47

=>36-x=47

=>x=-11

f: =>x(x-3)(11-x)(11+x)=0

hay \(x\in\left\{0;3;11;-11\right\}\)

g: =>-62-38-x+2x=-100

=>x-100=-100

hay x=0