a) 612 - 256 : X = 589

             256 : X = 612 - 589

             256 : X =      23

                      X = 256 : 23

                      X =  256/23

còn 2 câu nữa giúp luôn đi mà bn

bài này mình biết làm nhưng thôi

\(\frac{x+1}{2}=\frac{8}{x+1}\Rightarrow\left(x+1\right)^2=8.2\)

=>(x+1)²=16

=>x+1=\(\sqrt{16}\)

=>x+1=4

=>x=3

\(x\times6,4=14,3-8+38,6\\ x\times6,4=44,9\\ x=44,9:6,4\\ x=7,015625\)

\(92,6-\left(x+25,8\right)=37,04\\ x+25,8=92,6-37,04\\ x+25,8=55,56\\ x=55,56-25,8\\ x=29,76\)

`x-1/8=1/4xx2/3`

`x-1/8=1/6`

`x=1/6+1/8`

`x=7/24`

x - 1/8 = 1/6

x         = 6/48 + 8/48
x         = 14/48

x - 1/8 = 1/4 x 2/3

x - 1/8=   1/6

x        = 1/6 + 1/8

x       =  14/48

\(X:\dfrac{3}{9}=\dfrac{7}{8}\)

\(X=\dfrac{7}{8}\times\dfrac{3}{9}\)

\(X=\dfrac{21}{72}\)

\(\left(X-21\times13\right):11=30\)

\(X-273=330\)

\(X=603\)

`x:3/9=7/8`

`x=7/8xx3/9`

`x=7/24`

`-------------`

`(x-21xx13):11=30`

`(x-273):11=30`

`x-273=30xx11`

`x-273=330`

`x=330-273`

`x=57`

x = 4/5 : 3/4

x = 16/15

x = 5/9 : 8/3

x = 5/9 x 3/8

x = 5/24

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

\(x=\dfrac{16}{15}\)

-----------------------

\(x=\dfrac{5}{9}\times\dfrac{3}{8}\)

\(x=\dfrac{5}{24}\)

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

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

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

\(x=\dfrac{16}{15}\)

\(\dfrac{5}{9}:x=\dfrac{8}{3}\)

      \(x=\dfrac{5}{9}:\dfrac{8}{3}\)

      \(x=\dfrac{5}{9}\times\dfrac{3}{8}\)

     \(x=\dfrac{15}{72}\)

    \(x=\dfrac{5}{24}\)

\(\Leftrightarrow\left(x+2\right)^2\cdot\left(x+1\right)=0\)

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

\(\Leftrightarrow x^3+5x^2+8x+4=0\\ \Leftrightarrow\left(x+1\right)\left(x+2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)

Bài 2: 

Ta có: \(x\left(x-4\right)-x^2+8=0\)

\(\Leftrightarrow x^2-4x-x^2+8=0\)

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

hay x=2

1)=x²-49-x²

=-49

2)=>x²-4x-x²+8=0

=>-4x+8=0

=>-4x=-8

=>x=2

a: \(\Leftrightarrow x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)

hay \(x\in\left\{0;\sqrt{3};-\sqrt{3}\right\}\)

b: \(=\dfrac{x^3-3x^2+6x-8}{x-2}=\dfrac{x^2-2x-x^2+2x+4x-8}{x-2}=x^2-x+4\)