giúp mk vs nhé;tìm x biết (x+5)^2-(x-2)(x+3)=1 3(2x-1)(3x-1)-(2x-3)(9x-1)-3=-3 5(2x-1)-4(8-3x)=-5

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1. +)$(x+5)^2 – (x-2)(x+3) = 1$
   ⇔ $x^2 + 10x  + 25 – x^2 – x + 6 – 1 =0$
   ⇔ $9x + 30 = 0$
   ⇔ $9x = -30$
   ⇔ $x = \dfrac{-10}{3}$

   +)$3(2x -1)(3x-1) – (2x-3)(9x-1) – 3  = – 3$
   ⇔ $3(6x^2 -5x + 1) – (18x^2 – 29x + 3) – 3 + 3 = 0$

   ⇔ $18x^2 – 15x – 3 – 18x^2 + 29x – 3 = 0$

   ⇔ $14x – 6 = 0$

   ⇔ $14x = 6$

   ⇔ $x= \dfrac{3}{7}$

   +)$5(2x-1) – 4(8-3x) = -5$
   ⇔ $10x – 5 – 32+ 12x = -5$

   ⇔ $22x = 32$

   ⇔ $x = \dfrac{16}{11}$

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2. Đáp án:

    

   Giải thích các bước giải:

   $(x+5)^2-(x-2)(x+3)=1$

   $⇔x^2+10x+25-(x^2+3x-2x-6)=1$

   $⇔x^2+10x+25-x^2-3x+2x+6=1$

   $⇔(x^2-x^2)+(10x-3x+2x)=1-6-25$

   $⇔9x=-30$

   $⇔x=\dfrac{-10}{3}$

   $3(2x-1)(3x-1)-(2x-3)(9x-1)-3=-3$

   $⇔(6x-3)(3x-1)-(18x^2-2x-27x+3)=0$

   $⇔18x^2-6x-9x+3-18x^2+2x+27x-3=0$

   $⇔14x=0$

   $⇔x=0$

   $5(2x-1)-4(8-3x)=-5$

   $⇔10x-5-32+12x=-5$

   $⇔22x=-5+5+32$

   $⇔22x=32$

   $⇔x=\dfrac{16}{11}$

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