a; \(\dfrac{93}{17}\): \(x\) + (- \(\dfrac{21}{17}\)) : \(x\) + \(\dfrac{22}{7}\): \(\dfrac{22}{3}\) = \(\dfrac{5}{14}\)

   \(\dfrac{94}{17}\) \(\times\) \(\dfrac{1}{x}\) - \(\dfrac{21}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)

    \(\dfrac{72}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)

     \(\dfrac{72}{17x}\)        = \(\dfrac{5}{14}\) - \(\dfrac{3}{7}\)

      \(\dfrac{72}{17x}\)      = - \(\dfrac{1}{14}\)

      17\(x\)       = 72.(-14)

       17\(x\)     = - 1008

         \(x\)       = - 1008 : 17

         \(x\)       = - \(\dfrac{1008}{17}\)

Vậy \(x\) \(=-\dfrac{1008}{17}\)

b; - \(\dfrac{32}{27}\) - (3\(x\) - \(\dfrac{7}{9}\))³ = - \(\dfrac{24}{27}\)

          - \(\dfrac{32}{27}\)  + \(\dfrac{24}{27}\) = (3\(x\) - \(\dfrac{7}{9}\))³ 

          (3\(x-\dfrac{7}{9}\))³ = - \(\dfrac{8}{27}\)

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

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

           3\(x\)        = - \(\dfrac{2}{3}\) + \(\dfrac{7}{9}\)
           3\(x\)        = \(\dfrac{1}{9}\)

             \(x\)        = \(\dfrac{1}{9}\) : 3

             \(x\)       = \(\dfrac{1}{27}\)

Vậy \(x=\dfrac{1}{27}\)

dmm

a: \(\left(x+1\right)^3+\left(x-2\right)^3=2x^3+2\left(2x-1\right)^2-9\)

\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=2x^3+2\left(4x^2-4x+1\right)-9\)

\(\Leftrightarrow2x^3-3x^2+15x-7=2x^3+8x^2-8x-7\)

\(\Leftrightarrow-11x^2+23x=0\)

\(\Leftrightarrow x\left(-11x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{11}\end{matrix}\right.\)

1) \(\left(x-3\right)^2-4=0\)

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

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

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

2) \(x^2-2x=24\)

\(\Leftrightarrow x^2-2x-24=0\)

\(\Leftrightarrow x^2+4x-6x-24=0\)

\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Câu 3 số xấu rồi e

alo =0 nhé anh pop pop

n: ĐKXĐ: x<>0

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

=>\(\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right)-\left(x+\dfrac{1}{x}\right)+2=0\)

=>\(\left(x+\dfrac{1}{x}-2\right)\left(x+\dfrac{1}{x}-1\right)=0\)

=>\(\dfrac{x^2+1-2x}{x}\cdot\dfrac{x^2+1-x}{x}=0\)

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

=>\(\left(x-1\right)^2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0\)

=>\(\left(x-1\right)^2=0\)

=>x-1=0

=>x=1

p: \(x^4-4x^3+6x^2-4x+1=0\)

=>\(x^4-x^3-3x^3+3x^2+3x^2-3x-x+1=0\)

=>\(x^3\left(x-1\right)-3x^2\left(x-1\right)+3x\left(x-1\right)-\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x^3-3x^2+3x-1\right)=0\)

=>\(\left(x-1\right)^4=0\)

=>x-1=0

=>x=1

` 8/23 . 46/24 =1/3 .x`

`=>8/23 . 23/12 =1/3 . x`

`=> 1/3 . x=2/3`

`=>x=2/3 : 1/3`

`=>x=2/3 . 3`

`=> x= 6/3`

`=>x=2`

`----`

`1/5 : x= 1/5-1/7`

`=>1/5 : x=  7/35 - 5/35`

`=> 1/5 :x= 2/35`

`=>x= 1/5 : 2/35`

`=>x=1/5 . 35/2`

`=>x=7/2`

`----`

`4/9 - (x-1/2)^2 =1/3`

`=> (x-1/2)^2 =4/9-1/3`

`=> (x-1/2)^2 =4/9- 3/9`

`=> (x-1/2)^2 =1/9`

`=> (x-1/2)^2 = (+- 1/3)^2`

`@ TH1`

`x-1/2=1/3`

`=>x=1/3+1/2`

`=>x= 2/6 + 3/6`

``=>x= 5/6`

`@ TH2`

`x-1/2=-1/3`

`=>x=-1/3 +1/2`

`=>x= -2/6 + 3/6`

`=>x=1/6`

`----`

`3,2 . x-(4/5+2/3) : 3 2/3 = 7/10`

`=> 3,2 . x-22/15 : 11/3 = 7/10`

`=>  3,2 . x-22/15 = 7/10 . 11/3`

`=>  3,2 . x-22/15 =77/30`

`=> 3,2 .x= 77/30 + 22/15`

`=> 3,2 .x=121/30`

`=>x= 121/30. 5/16`

`=>x= 121/96`