Question

Câu 2.(1,5 điểm) Tìm x, biết:

a)     5x(x² – 9) = 0.                 b) 3(x+3) - x² - 3x  =0.                  c) x² – 9x – 10 = 0

Answer 1

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

Answer 2

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}

 

Answer 3

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

                

Answer 4

a) x ϵ {0;3;-3}

b) x ϵ {3;-3}

c) x ϵ {-1;10}