a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)

b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))

\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)

\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)

d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)

\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)

\(\Leftrightarrow2x^2+2x=2x^2+1\)

\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).

\(a,\Leftrightarrow x^3-8-x\left(x^2-9\right)=1\\ \Leftrightarrow x^3-8-x^3+9x=1\\ \Leftrightarrow9x=9\Leftrightarrow x=1\\ b,\Leftrightarrow8x^3+12x^2+6x+1-8x^3 +12x^2-6x+1-24x^2+24x-1=0\Leftrightarrow1=0\Leftrightarrow x\in\varnothing\)

a) \(\Leftrightarrow x^3-8-x^3+9x=1\)

\(\Leftrightarrow9x=9\Leftrightarrow x=1\)

b) \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2+24x-6=5\)

\(\Leftrightarrow24x=9\Leftrightarrow x=\dfrac{3}{8}\)

\((x+2)(x^2-2x+4)=(x-1)^3+3(x+1)^2\\\Leftrightarrow x^3+2^3=x^3-3x^2+3x-1+3\cdot(x^2+2x+1)\\\Leftrightarrow x^3 +8=x^3-3x^2+3x-1+3x^2+6x+3\\\Leftrightarrow x^3-x^3 +3x^2-3x-3x^2-6x=-1+3-8\\\Leftrightarrow -9x=-6\\\Leftrightarrow x=\dfrac{2}{3}\)

Vậy \(x=\dfrac{2}{3}\)

1: =>x^2+4x-21=0

=>(x+7)(x-3)=0

=>x=3 hoặc x=-7

2: =>(2x-5-4)(2x-5+4)=0

=>(2x-9)(2x-1)=0

=>x=9/2 hoặc x=1/2

3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15

=>-9x^2+27x+9x^2+18x+9=15

=>18x=15-9-27=-21

=>x=-7/6

6: =>4x^2+4x+1-4x^2-16x-16=9

=>-12x-15=9

=>-12x=24

=>x=-2

7: =>x^2+6x+9-x^2-4x+32=1

=>2x+41=1

=>2x=-40

=>x=-20

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

\(\Leftrightarrow x^3-3x^2+3x-1+3x^2+6x+1=x^3+8\)

\(\Leftrightarrow x^3+9x=x^3+8\)

hay \(x=\dfrac{8}{9}\)

\(-2x-\frac{3}{4}=x-\frac{3}{5}\)

\(3x=\frac{3}{5}-\frac{3}{4}=-\frac{3}{20}\)

\(x=-\frac{1}{20}\)

b) \(\left|\frac{x}{2}-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)

=> \(\orbr{\begin{cases}\frac{x}{2}-\frac{1}{3}=\frac{7}{4}\\\frac{x}{2}-\frac{1}{3}=-\frac{7}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{25}{6}\\x=-\frac{17}{6}\end{cases}}}\)

a, \(-2x-\frac{3}{4}=x-\frac{3}{5}\)

\(-2x-\frac{3}{4}-x+\frac{3}{5}=0\)

\(-3x-\frac{3}{20}=0\)

\(\frac{3}{20}=-3x\Leftrightarrow x=\frac{1}{20}\)

\(a,-2x+\frac{-3}{4}=x-\frac{3}{5}\)

\(-2x-x=\frac{3}{4}-\frac{3}{5}\)

\(-3x=\frac{15}{20}-\frac{12}{20}\)

\(-3x=\frac{3}{20}\)

\(x=\frac{3}{20}\div\left(-3\right)\)

\(x=\frac{3}{20}.\frac{-1}{3}\)

\(x=-\frac{1}{20}\)

Vậy \(x=-\frac{1}{20}\).

\(b,\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)

\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}\)

\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{6}{4}\)

\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{4}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=\frac{25}{12}\\\frac{1}{2}x=-\frac{17}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{25}{6}\\x=-\frac{17}{6}\end{cases}}\)

Vậy \(x\in\left\{\frac{25}{6};-\frac{17}{6}\right\}\).