\(9\times\left(\dfrac{-1}{3}\right)^3+\dfrac{1}{3}:0,25-5\dfrac{3}{4}=\dfrac{9}{1}\times\left(\dfrac{-1}{27}\right)+\dfrac{1}{3}:\dfrac{1}{4}-\dfrac{23}{4}\)

\(=\left(\dfrac{-9}{27}\right)+\dfrac{4}{3}-\dfrac{23}{4}=\left(\dfrac{-1}{3}\right)+\dfrac{4}{3}-\dfrac{23}{4}=\dfrac{3}{3}-\dfrac{23}{4}=1-\dfrac{23}{4}=\dfrac{4}{4}-\dfrac{23}{4}=\dfrac{-19}{4}\)

= -\(\dfrac{19}{4}\)

=-19/4

a,=11/12 x 8/15 - 1/6:2

  = 22/45-1/2

  = 73/180

b, =9/4:0.01 

   = 225

K nha

0,25 x 40

= 10

....

Cái chỗ phần b là ( 2 và 1/4 - 0,25 ) :0,01 phần 0,25x 40 nhé mn

a: \(0.4\cdot\sqrt{0.25-\sqrt{\dfrac{1}{4}}}=0.4\cdot\sqrt{0.25-0.5}\)(đề này sai rồi bạn)

b: \(\dfrac{3}{2}+2\left(x-1\right)=-5\dfrac{1}{2}\)

\(\Leftrightarrow2\left(x-1\right)=\dfrac{-11}{2}-\dfrac{3}{2}=-7\)

\(\Leftrightarrow x-1=\dfrac{-7}{2}\)

hay \(x=-\dfrac{5}{2}\)

\(a,\dfrac{-3}{4}x+1=\dfrac{5}{6}\\ \Rightarrow\dfrac{-3}{4}x=\dfrac{-1}{6}\\ \Rightarrow x=\dfrac{2}{9}\\ b,\left(2x-3\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=5\end{matrix}\right.\\ c,\dfrac{1}{2}-\left|x+1\right|=0,25\\ \Rightarrow\left|x+1\right|=0,25\\ \Rightarrow\left[{}\begin{matrix}x+1=0,25\\x+1=-0,25\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-0,75\\x=-1,25\end{matrix}\right.\)

a: =>-3/4x=-1/6

hay x=2/9

b: =>2x-3=0 hoặc x-5=0

hay x=3/2 hoặc x=5

c: =>|x+1|=1/4

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

hay \(x\in\left\{-\dfrac{3}{4};-\dfrac{5}{4}\right\}\)

b, \(=>2x-3=0\)      hay       \(x-5=0\)

           \(2x=0+3\)                   \(x=0+5\)

          \(2x=3\)                           \(x=5\)

            \(x=\dfrac{3}{2}\)

Vậy x ϵ { \(\dfrac{3}{2}\);5}

1, (0,25 - \(x\)) : - \(\dfrac{3}{5}\) = - \(\dfrac{3}{4}\)

    0,25 - \(x\)           = - \(\dfrac{3}{4}\) x (- \(\dfrac{3}{5}\))

     0,25 - \(x\)           = \(\dfrac{9}{20}\)

                \(x\)          = 0,25 - 0,45 

                \(x\)          = - 0,2 

2, - \(\dfrac{3}{8}\)\(x\) - 0,75 = - 1\(\dfrac{1}{2}\)

    - \(\dfrac{3}{8}\)\(x\) - 0,75  = -1,5

       \(\dfrac{3}{8}\)\(x\)             = - 0,75 + 1,5

       \(\dfrac{3}{8}\)\(x\)            =  0,75

          \(x\)            = 0,75 : \(\dfrac{3}{8}\)

          \(x\)           =   2

3, \(\dfrac{3}{5}\)\(x\) + \(\dfrac{2}{3}\) = 2\(\dfrac{2}{3}\)

     \(\dfrac{3}{5}\)\(x\) + \(\dfrac{2}{3}\) = \(\dfrac{8}{3}\)

     \(\dfrac{3}{5}\)\(x\)        = \(\dfrac{8}{3}\) - \(\dfrac{2}{3}\)

      \(\dfrac{3}{5}\)\(x\)       = 2

         \(x\)        = 2 : \(\dfrac{3}{5}\)

          \(x\)       = \(\dfrac{10}{3}\)

Đợi tí             

Nhấn đúng cho mình níc kia đi mình làm cho

X : 0,25 + X - 5 = 7

X x 4 + X - 5 = 7

X . ( 4 + 1 ) = 7 + 5

X . 5 = 12

X = 12 : 5 = \(\frac{12}{5}\)

X : 0,25 - X + 5 = 3

 X x 4 - X + 5 = 3

X x ( 4 - 1 ) = 3 - 5

X x 3 = -2

X = -2 : 3 = \(\frac{-2}{3}\)

1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)

\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)

\(\left|4-2x\right|=\dfrac{1}{3}.3\)

\(\left|4-2x\right|=1\)

=>\(4-2x=\pm1\)

+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)

\(2x=4-1\) \(2x=4-\left(-1\right)\)

\(2x=3\) \(2x=4+1\)

\(x=3:2\) \(2x=5\)

\(x=1,5\) \(x=5:2\)

Vậy x=1,5 \(x=2,5\)

Vậy x=2,5

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

\(9:\left|x+\left(-1\right)\right|=-3\)

\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)

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

=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )

Vậy x = \(\varnothing\)

3) \(\left|\dfrac{3}{2}-x\right|+\left(-\dfrac{4}{5}\right)=-\dfrac{14}{5}\)

\(\left|\dfrac{3}{2}-x\right|=\left(-\dfrac{14}{5}\right)-\left(-\dfrac{4}{5}\right)\)

\(\left|\dfrac{3}{2}-x\right|=\left(-\dfrac{14}{5}\right)+\dfrac{4}{5}\)

\(\left|\dfrac{3}{2}-x\right|=-\dfrac{10}{5}\)

\(\left|\dfrac{3}{2}-x\right|=-2\)

=> \(\dfrac{3}{2}-x\) không có giá trị nào thỏa mãn ( Vì giá trị tuyệt đối không thể là số nguyên âm )

Vậy x=\(\varnothing\)