Toán Lớp 8: tìm x: a, 3(2x-1) . (3x+1) + (2x-3) . (9x-1)= 0 b, (x+2) . (x ² – 2x +4 ) – x(x ² +2 ) = 5 c, 3(x+2) ² + (2x – 1) ² -7(x+3)(x+3) =

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1. Giải đáp + Lời giải và giải thích chi tiết:

   a) 3(2x-1)(3x+1)+(2x-3)(9x-1)=0

   <=>3(6x^2+2x-3x-1)+18x^2-2x-27x+3=0

   <=>18x^2+6x-9x-3+18x^2-2x-27x+3=0

   <=>36x^2-32x=0

   <=> 4x(9x – 8) = 0

   <=>\(\left[ \begin{array}{l}4x=0\\9x-8=0\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=0\\x=\dfrac{8}{9}\end{array} \right.\)

   Vậy S={0;8/9}

   b) (x+2)(x^2-2x+4)-x(x^2+2)=5

   <=>x^3+8-x^3-2x=5

   <=> 8-2x=5

   <=>2x=3

   <=>x=3/2

   Vậy S={3/2}

   c) 3(x+2)^2+(2x-1)^2-7(x+3)(x+3)=36

   <=>3(x^2+4x+4)+(4x^2-4x+1)-7(x^2+6x+9)=36

   <=>3x^2+12x+12+4x^2-4x+1-7x^2-42x-63=36

   <=>-34x-50=36

   <=>-34x=86

   <=>x=-86/34=-43/17

   Vậy S={-43/17}

   d) (x+3)(x+2)-(x-2)(x+5)=6

   <=>x^2+5x+6-(x^2+5x-2x-10)=6

   <=>x^2+5x+6-x^2-5x+2x+10=6

   <=>2x+16=6

   <=>2x=-10

   <=>x=-5

   Vậy S={-5}

   e) 3(2x-1)(3x-1)-(2x-3)(9x-1)=0

   <=>3(6x^2-2x-3x+1)-(18x^2-2x-27x+3)=0

   <=>18x^2-6x-9x+3-18x^2+2x+27x-3=0

   <=>14x=0<=>x=0

   Vậy S={0}.

    

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