\(a,5x\left(x^2-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,3\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow3\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ c,x^2-9x-10=0\\ \Leftrightarrow x^2+x-10x-10=0\\ \Leftrightarrow x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)

a, 5\(x\)(\(x^2\) - 9) = 0

    \(\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)

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

Vậy \(x\) \(\in\) { -3; 0; 3}

b, 3.(\(x+3\)) - \(x^2\) - 3\(x\) = 0

    3.(\(x+3\)) - \(x\).( \(x\) + 3) = 0

    (\(x+3\))( 3 - \(x\)) = 0

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

Vậy \(x\) \(\in\){ -3; 3}

c, \(x^2\) - 9\(x\) - 10 = 0

   \(x^2\) + \(x\) - 10\(x\)  - 10 = 0

   \(x.\left(x+1\right)\) - 10.( \(x-1\)) = 0

        (\(x+1\))(\(x-10\)) = 0

         \(\left[{}\begin{matrix}x+1=0\\x-10=0\end{matrix}\right.\)

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

Vậy \(x\) \(\in\){ -1; 10}

a) 5x(x2-9)=0

=> TH1 5x=0

      <=> x= 0

     TH2: 2x-9=0

       <=> 2x=9

       <=> x = \(\dfrac{9}{2}\)

b, 3(x+3) - x²- 3x = 0   

<=> 3x + 9 - x² -3x = 0

<=> - x² +9 = 0

<=> - x² = -9

 <=> x = 3

c, x² -9x -10 = 0

<=> x² -x + 10x -10 = 0

<=> x(x-1)+10(x-1)=0     

 <=> (x-1)(x+10)=0

=> TH1: x-1=0

       <=> x=1

      TH2: x +10=0

      <=>  x=-10

a có x=4 và b có x=5 bạn nhé

\(a,\left(x-4\right)\left(x+7\right)=0\Rightarrow\orbr{\begin{cases}x-4=0\\x+7=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-7\end{cases}}\)

\(b,x^2-5x=0\Rightarrow x\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)

a) x=10

b) x=1

\(\left(x+10\right)\cdot2=40\\ \Leftrightarrow x+10=20\\ \Leftrightarrow x=10\\ 63:\left(5x+4\right)=2^3-1=7\\ \Leftrightarrow5x+4=9\\ \Leftrightarrow5x=5\\ \Leftrightarrow x=1\)

\(x-4=26\\ \Leftrightarrow x=30\)

\(\left(40+x\right).2=180\\ \Leftrightarrow40+x=90\\ \Leftrightarrow x=50\)

\(\left(14+5x\right):2=3^5:3^2\\ \Leftrightarrow\left(14+5x\right):2=27\\ \Leftrightarrow14+5x=54\\ \Leftrightarrow5x=40\\ \Leftrightarrow x=8\)

a) x-4=26

x=26+4= 30

b) (40+x)x2=180

40+x=180:2

40+x= 90 

x=90-40=50

c) ( 14+5x):2=3⁵: 3²

14+5x:2=3³=27

14+5x=27x2

14+5x=54

5x=54-14

5x=40

x=8

x-4=26                    (40+x).2=180               (14+5.x):2=3⁵:3²

  x=26+4                  40+x= 90                      (14+5.x):2=3³

  x=30                            x=90-40                  (14+5.x):2=27

                                     x=50                          (14+5.x)=27.2

                                                                       14+5.x=54

                                                                           5.x=54-14

                                                                           5.x=40

                                                                              x=40:5

                                                                             x=8

vừa hãy thấy gửi rồi mà
Spam à:D?

cái này tớ làm rồi mà

a) x-4=26

x=26+4= 30

b) (40+x)x2=180

40+x=180:2

40+x= 90 

x=90-40=50

c) ( 14+5x):2=3⁵: 3²

14+5x:2=3³=27

14+5x=27x2

14+5x=54

5x=54-14

5x=40

x=8

a: \(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

b: \(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

a) Ta có: \(x^2-2x+1=25\)

\(\Leftrightarrow\left(x-1\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

b) Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x=20\)

hay x=2

c) Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

d) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

a,\(< =>\left(x-1\right)^2-5^2=0< =>\left(x-1-5\right)\left(x-1+5\right)=0\)

\(< =>\left(x-6\right)\left(x+4\right)=0=>\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

b,\(< =>25x^2+10x+1-25x^2+9-30=0\)

\(< =>10x-20=0< =>10\left(x-2\right)=0< =>x=2\)

c,\(< =>x^3-1-x\left(x^2-4\right)-5=0\)

\(< =>x^3-1-x^2+4x-5=0< =>4x-6=0< =>x=\dfrac{6}{4}\)\(d,< =>\left(x-2\right)^3-x^3+3^3+6x^2+12x+6-15=0\)

\(< =>x^3-6x^2+12x-x^3+6x^2+12x+10=0\)

\(< =>24x+10=0< =>x=-\dfrac{5}{12}\)

a: Ta có: \(x^2-2x+1=25\)

\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)

b: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x=20\)

hay x=2

c: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

n = 2⁵.3^{x^2}

Vì n có 60 ước nên ta có:

(5 + 1)(x² + 1) = 60

=> 6(x² + 1) = 60

=> x² + 1 = 10

=> x² = 9

=> x = 3 (Vì x thuộc N*)

Do đó n = 2⁵.3^{3^2} = 2⁵.3⁹ = 629856

629856 , ủng hộ mk nha

n=2592 mới đúng chứ