Tìm x, bik a) 25x^2-2=0 b) (x+2)^2=(x-2)^2 c) (x+2)^2-x+4=0

1 bình luận về “Tìm x, bik a) 25x^2-2=0 b) (x+2)^2=(x-2)^2 c) (x+2)^2-x+4=0”

1. Giải đáp:

    

   Lời giải và giải thích chi tiết:

   a) 25x^2 – 2 = 0

   => ( 5x )^2 – \sqrt{2}^2 = 0

   => ( 5x – \sqrt{2} )( 5x + \sqrt{2} ) = 0

   => $[\begin{array}{c}5x–\sqrt{2}=0\\ 5x+\sqrt{2}=0\end{array}$ => $[\begin{array}{c}x={\displaystyle \frac{\sqrt{2}}{5}}\\ x={\displaystyle \frac{\sqrt{-2}}{5}}\end{array}$

   Vậy x in { +-\dfrac{\sqrt{2}}{5} }

   b) ( x + 2 )^2 = ( x – 2 )^2

   => ( x + 2 )^2 – ( x – 2 )^2 = 0

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

   => 4. 2x = 0

   => 8x = 0

   => x = 0

   Vậy x = 0

   c) ( x + 2 )^2 – x + 4 = 0

   => x^2 + 4x + 4 – x + 4 = 0

   => x^2 + 3x + 8 = 0

   => x^2 + 2 . x . 3/2 + ( 3/2 )^2 – ( 3/2 )^2 + 8 = 0

   => ( x + 3/2 )^2 + 23/4 = 0

   => ( x + 3/2 )^2 = -23/4 ( Vô lý $)$

   => x in ∅

   Trả lời