x+1526=6924

        x=6924-1526

        x=5398

x*9=2763

   x=2763/9

   x=307

x-636=5616

      x=5616+636

      x=6252

x/3=1628

   x=1628*3

   x=4884

đúng thì tick cho mình nhé

a, x+1536= 6924 

     x         = 6924 - 1536

     x         = 5388

b, x-636=5618 

    x       = 5618 + 636

    x      =   6254

c, x:3=1628

    x   = 1628 x 3

    x   = 4884

a x + 1536 = 6924

           x     = 6924 - 1536

           x     = 5388

B, x - 636 = 5618

           x    = 5618 + 636

           x    = 6254

C, x : 3= 1628

          x= 1628 x 3

          x=4884

mình cảm ơn 

Tìm x

x : 3 = 1628

x = 1628 x 3

x= 4884

x : 3 = 1628

x = 1628 x 3

x= 4884

a: \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow x^2+7x-x^2-x+6=0\)

hay x=-1

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

\(\Leftrightarrow x+2=0\)

hay x=-2

b. (x + 2)² - x² + 4 = 0

<=> (x + 2 - x)(x + 2 + x) + 4 = 0

<=> 2(2 + 2x) + 4 = 0

<=> 4(1 + x) + 4 = 0

<=> 4(1 + x) = -4

<=> 1 + x = -1

<=> x = -1 - 1

<=> x = -2

\(a,\) \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)\)

\(\Leftrightarrow x^2+7x-x^2-3x+2x+6\\ \Leftrightarrow6x=0\\ \Leftrightarrow x=0\)

    \(Vậy...\)

\(b,\) \(\left(x+2\right)^2-\left(x^2-4\right)=0\)

\(\Leftrightarrow x^2+4x+4-x^2+4=0\\ \Leftrightarrow4x+8=0\\ \Leftrightarrow x=-2\)

a) x * 2 + x * 3 = 60

x * ( 2 + 3 ) = 60

x = 60 : 5

x = 12

b) x * 15 - x * 9 = 78

x * ( 15 - 9 ) = 78

x = 78 : 6

x = 13

\(a,\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\\ \Rightarrow2x=-4\\ \Rightarrow x=-2\\ b,\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\\ \Rightarrow\left(2021x-1\right)\left(x-2020\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2020=0\\2021x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)

a) \(\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\)

\(\Rightarrow2x=-4\Rightarrow x=-2\)

b) \(\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\Rightarrow\left(x-2020\right)\left(2021x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)