a) \(\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\Leftrightarrow x=-\frac{6}{11}\)

b) \(\Leftrightarrow\left(\frac{43}{8}+x-\frac{173}{24}\right).-\frac{3}{50}=1\Leftrightarrow\frac{-11}{6}+x=-\frac{50}{3}\Leftrightarrow x=-\frac{37}{2}\)

Đề sai rồi bạn!!!
 

?????

\(=>1-\left(\frac{43}{8}+x-\frac{173}{24}\right)=0\)

\(=>\frac{43}{8}+x-\frac{173}{24}=1\)

\(=>\frac{43}{8}+x=\frac{197}{24}\)

\(=>x=\frac{17}{6}\)

=>  1 - (43/8+x-173/24) = 0

=>  43/8 + x - 173/24 = 1

=>  43/8 + x = 197/24

=>  x = 17/6

k nha

1-(43/8+x-173/24):50/3=0

a: x=2/3-4/5=10/15-12/15=-2/15

b: 1/2-x=7/12

=>x=1/2-7/12=-1/12

c: =>7/2:x=-7/2

=>x=-1

d: =>1/6x=3/8-5/2=3/8-20/8=-17/8

=>x=-17/8*6=-102/8=-51/4

e: =>1,5x=-1,5

=>x=-1

(x-11/6):50/3=2

x-11/6=100/3

x=211/6

\(2-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)

\(\left(x-\frac{11}{6}\right)\cdot\frac{3}{50}=2\)

\(\Rightarrow x-\frac{11}{6}=\frac{100}{3}\)

\(\Rightarrow x=\frac{211}{6}\)

\(2-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)

\(2-\left(x+\frac{11}{6}\right):\frac{50}{3}=0\)

\(2-\frac{3}{50}x-\frac{11}{100}=0\)

\(\frac{3}{50}x=\frac{189}{100}\)

\(x=\frac{63}{2}\)

a: Ta có: \(2x^3-18x=0\)

\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

b: Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)

\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)

\(\Leftrightarrow-13x=13\)

hay x=-1

c: Ta có: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)

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

\(\Leftrightarrow3x=12\)

hay x=4

a) 2x³-18x=0

⇔ 2x(x²-9)=0

⇔ 2x(x-3)(x+3)=0

⇔ \(\left\{{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

b)(3x-1)(2x+1)-6x(x+2)=11

⇔ 6x²+x-1-6x²-12x=11

⇔ -11x=12

\(\Leftrightarrow x=-\dfrac{12}{11}\)

c) (x-1)³-(x+2).(x²-2x+4)=3.(1-x²)

⇔ x³-3x²+3x-1-x³-8-3+3x²=0

⇔ 3x=12

⇔   x=4

c. (x - 1)³ - (x + 2)(x² - 2x + 4) = 3(1 - x²)

<=> (x³ - 3x² + 3x - 1) - (x³ - 2x² + 4x + 2x² - 4x + 8) = 3 - 3x²

<=> x³ - 3x² + 3x - 1 - x³ + 2x² - 4x - 2x² + 4x - 8 = 3 - 3x²

<=> x³ - x³ - 3x² + 2x² - 2x² + 3x² + 3x - 4x + 4x = 3 + 1 + 8

<=> 3x = 12

<=> x = 4

a.\(\dfrac{1}{3}\) + x  = \(\dfrac{5}{6}\)

       x = \(\dfrac{5}{6}\) - \(\dfrac{1}{3}\)

      x = \(\dfrac{1}{2}\)

b. | x-1| - \(\dfrac{2}{5}\) = \(\dfrac{11}{10}\) 

   | x-1|        = \(\dfrac{11}{10}\) + \(\dfrac{2}{5}\)

  |x-1|        = \(\dfrac{3}{2}\)

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

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

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

c, \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = 1

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

             \(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = \(\dfrac{2}{3}\)

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

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

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

                     \(x\) = -4

d, \(\dfrac{x+2}{3}\) = \(\dfrac{27}{x+2}\)

(x+2)² = 27.3

(x+2) =9²

\(\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)