=>x(x-2)=0

=>x=0 hoặc x=2

\(3x^4-24x=0\\ \Rightarrow3x\left(x^3-8\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x=0\\x^3-8=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^3=8\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)

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

\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)

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

a: \(x\cdot\dfrac{3}{4}+x=\dfrac{7}{8}\)

\(\Leftrightarrow x\cdot\dfrac{7}{4}=\dfrac{7}{8}\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Ta có : 

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

\(\Leftrightarrow\)\(4x^2-4x-3x^2+15=x-3-x-4\)

\(\Leftrightarrow\)\(x^2-4x+15=-7\)

\(\Leftrightarrow\)\(\left(x^2-2.x.2+2^2\right)+11=-7\)

\(\Leftrightarrow\)\(\left(x-2\right)^2=-18\)

Mà \(\left(x-2\right)^2\ge0\) \(\left(\forall x\inℝ\right)\)

\(\Rightarrow\)\(x\in\left\{\varnothing\right\}\)

Vậy không có giá trị nào của x thoã mãn đề bài 

Chúc bạn học tốt ~ 

Ta có: \(x^5-2x^4+3x^3-4x^2+2\)

\(=x^5-x^4-x^4+x^3+2x^3-2x^2-2x^2+2\)

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

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

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

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 \(x=\dfrac{5}{6}\)

2x-3=x+1/2

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

2x-x =1/2+3                           4x-x-1/2=2x+1/2-5                             d,(x+1/2).(x-3/4)=0

x=7/2                                4x-x-2x  =1/2-5+1/2                                 \(\orbr{\begin{cases}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{cases}}\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{4}\end{cases}}\)

                                            x=-4

e,(2x-1)(3x+1/5)=0

\(\orbr{\begin{cases}2x-1=0\\3x+\frac{1}{5}=0\end{cases}}\orbr{\begin{cases}2x=1\\3x=\frac{1}{5}\end{cases}}\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{15}\end{cases}}\)

f, 4x²-2x=0

Các câu mk chưa làm thì bạn cứ chờ để mk suy nghĩ.