Cách 1 : -0,52 : x = -9,36 : 16,83

-0,52 : x = -\(\frac{104}{187}\)

x = -0,25 : -\(\frac{104}{187}\)

x = \(\frac{187}{416}\)

Cách 2 : -0,52 : x = -9,36 : 16,83

=> \(\frac{-0.25}{x}\) = \(\frac{-9.36}{16.83}\)

=>x. (-9,36) = -0,25. 16,83

x.(-9,36) = -4,2075

x = -4,2075 : (-9,36)

x = \(\frac{187}{416}\)

\(\Rightarrow x=\frac{-0,52.16,83}{-9,86}=\frac{187}{200}\)

#Walker

\(-0,52:x=-9,36:16,83\)

\(\Rightarrow-0,52:x=-\frac{104}{187}\)

\(\Rightarrow-\frac{13}{25}:x=-\frac{104}{187}\)

\(\Rightarrow x=\frac{13}{25}:\frac{104}{187}\)

\(\Rightarrow x=\frac{187}{200}\)

Vậy \(x=\frac{187}{200}.\)

Chúc bạn học tốt!

a) (x-1)(x-2)=0

x-1=0 --> x = 1

x-2=0 --> x = 2

d) x^2(x + 1) + 27(x + 1)=0

(x+1)(x^2+27)=0

x+1=0 --> x = -1

x^2+27=0 (vô lí)

a) \(x\left(x-2\right)-x+2=0\)

\(x\left(x-2\right)-\left(x-2\right)=0\)

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

TH1:x-1=0⇒x=1

TH2:x-2=0⇒x=2

a) 

TH1:x-1=0⇒x=1

TH2:x-2=0⇒x=2

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

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

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

TH1:3-3x+1=0⇒x\(=\dfrac{4}{3}\)

TH2:3-x-1=0⇒x=2

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

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

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

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

\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)

\(\Leftrightarrow3x=3\)

hay x=1

Vậy: S={1}

b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)

\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)

\(\Leftrightarrow6x=-20\)

hay \(x=-\dfrac{10}{3}\)

c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)

\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)

\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)

\(\Leftrightarrow17x=17\)

hay x=1

a: \(\Leftrightarrow\left(x-1\right)^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)

\(-0.5x:x=-9.36:16.38\)

\(\Leftrightarrow-0.5x:x=-\frac{4}{7}\)

\(\Leftrightarrow-\frac{1}{2}x:x=-\frac{4}{7}\Leftrightarrow\left(-\frac{1}{2}:1\right)x=-\frac{4}{7}\)

\(\Leftrightarrow-\frac{1}{2}x=-\frac{4}{7}\Leftrightarrow x=\left(-\frac{4}{7}\right):\left(-\frac{1}{2}\right)\)

\(x=1\frac{1}{7}\Leftrightarrow x=\frac{8}{7}\)

xin lỗi tai mik lười nên mik viết nha pạn ...

- bạn tính phép tính về bên phải trước 

- rồi lấy -0,5 chia cho kết qur vừa tìm là ra x nha 

~ hok tốt ~