Tìm x, biết: (x – 7) (x + 2) = 0 x*2 – 6x = 0 x*4 – 3x*2 = 0 3x*2 – 2x = 0 4x*2 – 12x + 9 = 9 (2x – 2)*3 = 64

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1. Đáp án:

    

   Giải thích các bước giải:

   1, 

   `(x-7)(x+2)=0<=>`\(\left[ \begin{array}{l}x-7=0\\x+2=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=7\\x=-2\end{array} \right.\)

   Vậy `x \in {-2;7}`

   $\\$
   2,

   `x^2-6x=0<=>x(x-6)=0<=>`\(\left[ \begin{array}{l}x=0\\x-6=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=0\\x=6\end{array} \right.\)

   Vậy `x \in { 0;6}`

   $\\$
   3,

   `x^4-3x^2=x^2(x^2-3)=0<=>`\(\left[ \begin{array}{l}x^2=0\\x^2-3=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=0\\x= ±\sqrt3\end{array} \right.\) 

   Vậy `x \in {+-\sqrt{3};0}`

   $\\$

   4,

   `3x^2-2x=0<=>x(3x-2)=0<=>`\(\left[ \begin{array}{l}x=0\\3x-2=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=0\\x=\dfrac32\end{array} \right.\) 

   Vậy `x \in {0;3/2}`

   $\\$
   5,

   `4x^2-12x+9=9<=>4x^2-12x=0<=>4x(x-3)=0<=>`\(\left[ \begin{array}{l}4x=0\\x-3=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\) 

   Vậy `x \in {0;3}`

   $\\$
   6,

   `(2x-3)^3=64<=>(2x-3)^3=4^3<=>2x-3=4<=>2x=7<=>x=7/2`

   Vậy `x=7/2`

   Trả lời

2. Mình gửi bạn nhé:

           (x – 7) (x + 2) = 0

   <=>\(\left[ \begin{array}{l}x-7=0\\x+2=0\end{array} \right.\) 

   <=>\(\left[ \begin{array}{l}x=7\\x=-2\end{array} \right.\) 

   Vậy x=7 hoặc x=-2

           x*2 – 6x = 0

   <=>x(x-6)=0

   <=>\(\left[ \begin{array}{l}x=0\\x-6=0<=>x=6\end{array} \right.\) 

   Vậy x=0 hoặc x=6

          x*4 – 3x*2 = 0

   <=>x*2(x*2-3)=0

   <=>\(\left[ \begin{array}{l}x=0\\x^{2} =3 <=>x=+-căn 3\end{array} \right.\) 

   Vậy…….

         3x*2 – 2x = 0

   <=>x(3x-2)=0

   <=>\(\left[ \begin{array}{l}x=0\\3x=2<=>x=3/2\end{array} \right.\) 

   Vậy……..

          4x*2 – 12x + 9 = 9

   <=>4x*2-12x=0

   <=>x*2-3x=0

   <=>x(x-3)=0

   <=>x=0 hoặc x=3

    Vậy……………

         (2x – 2)*3 = 64

   <=>2x-2=8

   <=>x-1=4

   <=>x=5

   Vậy …….

   Cho xin ctlhn ạ

    

   Trả lời