(4/15-1/6)x = -3/4.8/27

=> 1/10x = -2/9

=> x = -20/9

Trả lời

(4/15-1/6)x=-3/4.8/27

        1/10x=-2/9

               x=-2/9:1/10

              x=-20/9

Trong bài có sai sót mong bạn bỏ qua !

\(\left(\frac{4}{15}-\frac{1}{6}\right).x=-\frac{3}{4}.\frac{8}{27}\)

\(\Leftrightarrow\frac{1}{10}x=-\frac{2}{9}\)

\(\Leftrightarrow x=-\frac{20}{9}\)

a) x = 5/8 : 3/4

x = 5/6

b) x = 5/6 - 1/12

x = 3/4

a)x=\(\dfrac{5}{8}:\dfrac{3}{4}=\dfrac{5}{6}\)

b) x=\(\dfrac{5}{6}-\dfrac{1}{12}=\dfrac{3}{4}\)

a \(x=\dfrac{5}{8}:\dfrac{3}{4}\)

\(x=\dfrac{5}{8}\times\dfrac{4}{3}\)

\(x=\dfrac{5}{6}\)

b \(x=\dfrac{5}{6}-\dfrac{1}{12}\)

\(x=\dfrac{10}{12}-\dfrac{1}{12}\)

\(x=\dfrac{9}{12}\)

\(x=\dfrac{3}{4}\)

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.\(\dfrac{1}{3}\) + 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: \(\Leftrightarrow\left(x-1\right)^2=81\)

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

a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)

\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)

\(\Leftrightarrow2x=-8\)

hay x=-4

b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)

\(\Leftrightarrow-10x=-10\)

hay x=1

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

\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)

\(\Leftrightarrow-4x=-8\)

hay x=2

a) Ta có: \(\left(2x-5\right)^3=216\)

\(\Leftrightarrow2x-5=6\)

\(\Leftrightarrow2x=11\)

hay \(x=\dfrac{11}{2}\)

b) Ta có: \(2x-3⋮x+4\)

\(\Leftrightarrow-11⋮x+4\)

\(\Leftrightarrow x+4\in\left\{1;-1;11;-11\right\}\)

hay \(x\in\left\{-3;-5;7;-15\right\}\)

Alo, sugeni two wai phem. Si ga no, you woo be the me that nas te, ai gi da

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

\(a,x+\frac{1}{2}=\frac{3}{4}.\frac{8}{6}\)

\(x+\frac{1}{2}=1\)

\(x=\frac{1}{2}\)

\(b,x-\left(\frac{1}{6}+\frac{1}{3}\right)=\frac{5}{2}.\frac{7}{5}\)

\(x-\frac{1}{6}-\frac{1}{3}=\frac{7}{2}\)

\(x-\frac{1}{6}=\frac{23}{6}\)

\(x=4\)

Câu c,d đâu

câu c là 1/8+7/8.x=13/8