tìm x biết a ) 5x^2 – 10x^2 = 0 b) 16x ( x -7 ) – x + 7 = 0 c) ( 3x – 2 ) ^2 ( 2x + 3 ) ^2 = 0 d) x^2 = 5x + 14

1 bình luận về “tìm x biết a ) 5x^2 – 10x^2 = 0 b) 16x ( x -7 ) – x + 7 = 0 c) ( 3x – 2 ) ^2 ( 2x + 3 ) ^2 = 0 d) x^2 = 5x + 14”

1. Lời giải:

   a, 5x^2 – 10x^2 = 0

     <=> x^2(5 – 10) = 0

     <=> -5x^2 = 0

     <=> x = 0

   Vậy x = 0

   b, 16x(x – 7) – x + 7 = 0

     <=> 16x(x – 7) – (x – 7) = 0

     <=> (x – 7)(16x – 1) = 0

     <=> $\left[\begin{matrix} x – 7 = 0\\ 16x – 1 = 0\end{matrix}\right.$

     <=> $\left[\begin{matrix} x = 7\\ x = \dfrac{1}{16}\end{matrix}\right.$

   Vậy x = 7 hoặc x = 1/16

   c, (3x – 2)^2 – (2x + 3)^2 = 0

     <=> [(3x – 2) – (2x + 3)][(3x – 2) + (2x + 3)] = 0

     <=> (3x – 2 – 2x – 3)(3x – 2 + 2x + 3) = 0

     <=> (x – 5)(5x + 1) = 0

     <=> $\left[\begin{matrix} x – 5 = 0\\ 5x + 1 = 0\end{matrix}\right.$

     <=> $\left[\begin{matrix} x = 5\\ x = \dfrac{-1}{5}\end{matrix}\right.$

   Vậy x = 5 hoặc x = (-1)/5

   d, x^2 = 5x + 14

     <=> x^2 – 5x – 14 = 0

     <=> x^2 + 2x – 7x – 14 = 0

     <=> x(x + 2) – 7(x + 2) = 0

     <=> (x + 2)(x – 7) = 0

     <=> $\left[\begin{matrix} x + 2 = 0\\ x – 7 = 0\end{matrix}\right.$

     <=> $\left[\begin{matrix} x = -2\\ x = 7\end{matrix}\right.$

   Vậy x = -2 hoặc x = 7

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