\(\dfrac{1}{2}-3x+\left|x-1\right|=0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}-0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}\\ \Rightarrow\left|x-1\right|=\dfrac{1}{2}-3x\\ \Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}-3x\\x-1=-\dfrac{1}{2}+3x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+3x=\dfrac{1}{2}+1\\x-3x=-\dfrac{1}{2}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=\dfrac{3}{2}\\2x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)

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\(\dfrac{1}{2}\left|2x-1\right|+\left|2x-1\right|=x+1\\ \Rightarrow\left|2x-1\right|\cdot\left(\dfrac{1}{2}+1\right)=x+1\\ \Rightarrow\left|2x-1\right|\cdot\dfrac{3}{2}=x+1\\ \Rightarrow\left|2x-1\right|=x+1:\dfrac{3}{2}\\ \Rightarrow\left|2x-1\right|=x+\dfrac{2}{3}\\ \Rightarrow\left[{}\begin{matrix}2x-1=x+\dfrac{2}{3}\\2x-1=-x-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=\dfrac{2}{3}+1\\2x+x=-\dfrac{2}{3}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\3x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{1}{9}\end{matrix}\right.\)

a: \(\dfrac{3x+2}{4}-\dfrac{3x+1}{3}=\dfrac{5}{6}\)

=>3(3x+2)-4(3x+1)=10

=>9x+6-12x-4=10

=>-3x+2=10

=>-3x=8

=>x=-8/3

b: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{9x-10}{4-x^2}\)

=>(x-1)(x-2)-x(x+2)=-9x+10

=>x^2-3x+2-x^2-2x=-9x+10

=>-5x+2=-9x+10

=>x=2(loại)

Đặt thừa số chung là được nhé bạn

-3x²+6x⁴=0

-3x²(1+2x²)=0

Suy ra : TH1 -3x²=0  => x²=0 => x=0

             TH2 

nhưng tại sao lại là ( 1+2x²) ạ ?

ĐKXĐ: ...

Đặt \(x^2-3x+2=t\Rightarrow2x^2-6x+1=2t-3\)

\(\frac{4}{t}-\frac{3}{2t-3}+1=0\)

\(\Leftrightarrow8t-12-3t+t\left(2t-3\right)=0\)

\(\Leftrightarrow2t^2+2t-12=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-3x+2=2\\x^2-3x+2=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-3x=0\\x^2-3x+5=0\end{matrix}\right.\)

a) \(\dfrac{1}{4}+\dfrac{3}{4}:x=-2\)

\(\dfrac{3}{4}:x=-2-\dfrac{1}{4}=\dfrac{-8}{4}-\dfrac{1}{4}\)

\(\dfrac{3}{4}:x=\dfrac{-9}{4}\)

\(x=\dfrac{3}{4}:\dfrac{-9}{4}=\dfrac{3}{4}.\dfrac{-4}{9}\)

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

b) \(\dfrac{3}{4}+2.\left(2x-\dfrac{2}{3}\right)=-2\)

\(2.\left(2x-\dfrac{2}{3}\right)=-2-\dfrac{3}{4}=\dfrac{-8}{4}-\dfrac{3}{4}\)

\(2.\left(2x-\dfrac{2}{3}\right)=\dfrac{-11}{4}\)

\(2x-\dfrac{2}{3}=\dfrac{-11}{4}:2=\dfrac{-11}{4}.\dfrac{1}{2}\)

\(2x-\dfrac{2}{3}=\dfrac{-11}{8}\)

\(2x=\dfrac{-11}{8}+\dfrac{2}{3}=\dfrac{-33}{24}+\dfrac{16}{24}\)

\(2x=\dfrac{-17}{24}\)

\(x=\dfrac{-17}{24}:2=\dfrac{-17}{24}.\dfrac{1}{2}\)

\(x=\dfrac{-17}{48}\)

c) \(\left(\dfrac{1}{2}+5x\right).\left(2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}+5x=0\\2x-3=0\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{10}\\x=\dfrac{3}{2}\end{matrix}\right.\)

a, 1/4 + 3/4 : x = -2

     3/4 : x = -2 - 1/4 

     3/4 : x = -9/4

             x = 3/4 : -9/4

             x = -1/3

b, 3/4 + 2 . ( 2x - 3 ) = 0 

             2 . ( 2x - 3 ) = 0 - 3/4

             2 . ( 2x - 3 ) = -3/4

                     2x - 3 = -3/4 : 2

                     2x - 3 = -3/8

                          2x = -3/8 + 3

                          2x = 21/8

                            x = 21/8 : 2

                            x = 21/16

Ta có: \(6x+1=2\left(3x-1\right)+3\)

Vì \(2\left(3x-1\right)⋮\left(3x-1\right)\Rightarrow3⋮\left(3x-1\right)\)

\(\Rightarrow3x-1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow3x=\left\{2;0;4;-2\right\}\)

\(\Rightarrow x=\left\{\frac{2}{3};0;\frac{4}{3};\frac{-2}{3}\right\}\)

Vì biểu thức là số nguyên 

Vậy x = 0

Đầu bài có thiếu không nhỉ?

\(6x+1=6x-2+3.\)

\(\left(6x-2\right)+3=2\left(3x-1\right)+3⋮3x-1\)

a: \(\left(x^2+x\right)^2+2\left(x^2+x\right)-8=0\)

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

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

hay \(x\in\left\{-2;1\right\}\)

b: \(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+24=0\)

\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x-12\right)+24=0\)

\(\Leftrightarrow\left(x^2+x\right)^2-14\left(x^2+x\right)+48=0\)

\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x-8\right)=0\)

hay \(x\in\left\{-3;2;\dfrac{-1+\sqrt{33}}{2};\dfrac{-1-\sqrt{33}}{2}\right\}\)