36 : x = 4

x = 36 : 4

x = 9

36 : x = 4

x = 36 : 4

x = 9

36  :  x  =  4                                                 x  +  12  =  37

         x  =  36 : 4                                          x            = 37 – 12

         x  =  9                                                 x            =  25

782  -   X   =  215                                                       36  : x   = 4

               X= 782 – 125                                                        x =36 : 4

                   X= 657                                                                              x = 9

a) x  x 4 = 36                                        b) 70 :  x  = 7

            x = 36 : 4                                             x = 70 : 7

                                x = 9                                                     x = 10

Đề bài :

\(\left(x-2\right)\left(x-4\right)\left(x+6\right)\left(x+8\right)=-36\)

\(x=+_-\sqrt{34}-2,\)

\(x=-3\sqrt{2}-2,\)

\(x=3\sqrt{2}-2\)

(-5)+(-11)

\(\left(x-1\right)\left(x-4\right)\left(x-5\right)\left(x-8\right)+36=0\)

\(\left[\left(x-1\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x-5\right)\right]+36=0\)

\(\left(x^2-9x+8\right)\left(x^2-9x+20\right)+36=0\)

Đặt \(a=x^2-9x+14\)ta có :

\(\left(a-6\right)\left(a+6\right)+36=0\)

\(a^2-6^2+36=0\)

\(a^2=0\)

Thay \(a=x^2-9x+14\)ta có :

\(\left(x^2-9x+14\right)^2=0\)

\(\Leftrightarrow x^2-9x+14=0\)

\(\Leftrightarrow x^2-2x-7x+14=0\)

\(\Leftrightarrow x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}}\)

Vậy,...........

\(\left(x-1\right)\left(x-4\right)\left(x-5\right)\left(x-8\right)+36=0\)

\(\left(x-1\right)\left(x-8\right)\left(x-4\right)\left(x-5\right)+36=0\)

\(\left(x^2-9x+8\right)\left(x^2-9x+20\right)+36=0\)

Đặt \(x^2-9x+14=t\)

\(\left(t-6\right)\left(t+6\right)+36=0\)

\(t^2-36+36=0\)

\(\Rightarrow t^2=0\)

\(\Rightarrow x^2-9x+16=0\)

\(\Rightarrow x^2-2.\frac{9}{2}x+\frac{81}{4}=\frac{17}{4}\)

\(\Rightarrow\left(x-\frac{9}{2}\right)^2=\frac{17}{4}\)

\(\Rightarrow x=\hept{\begin{cases}\sqrt{\frac{17}{4}}+\frac{9}{2}\\-\sqrt{\frac{17}{4}}+\frac{9}{2}\end{cases}}\)

\(x^3-4x^2-9x+36=0\)

\(x^2\left(x-4\right)-9\left(x-4\right)=0\)

\(\left(x-4\right)\left(x^2-9\right)=0\)\(\)

\(\Rightarrow x-4=0\) hay \(x^2-9=0\)

\(\Rightarrow x=4\) hay \(x^2=9=3^2\)

\(\Rightarrow x=4\) hay \(x=\pm3\)

⇔x²(x-4) -9(x-4) = 0

⇔(x-4).(x-3).(x+3) = 0

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

Lớp 4?

a) \(\dfrac{x+1}{4}=\dfrac{36}{x+1}\)

\(\Rightarrow\left(x+1\right)^2=144\)

\(\Rightarrow\left[{}\begin{matrix}x+1=12\\x+1=-12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=11\\x=-13\end{matrix}\right.\)

Vậy: \(x\in\left\{11;-13\right\}\)

b) \(\dfrac{x}{7}=\dfrac{55x-4}{28}\)

\(\Rightarrow4x=55x-4\)

\(\Rightarrow-51x=-4\)

\(\Rightarrow x=\dfrac{4}{51}\)

Vậy: \(x=\dfrac{4}{51}\)

a) \(\dfrac{x + 1}{4} = \dfrac{36}{x + 1} \) 

\(\Rightarrow\) \(( x + 1 )( x + 1 ) = 36 . 4 \)  

\(\Rightarrow ( x + 1 )^2 = 144 \) 

\(\Rightarrow ( x + 1 )^2 = 12^2 = ( -12 )^2 \) 

\(\Rightarrow\) \(x + 1 ∈ \) { \(12 ; -12 \) }

\(\Rightarrow \) \(x \) \(∈ \) { \(11 ; -13 \) }

Vậy \(x ∈ \) { \(11 ; -13 \) }