(x+2)(x+3)-(x-2)(x-5)=-4 (x+1)( $x^{2}$ -x+1)-x(x+3)(x-3)=8

0 bình luận về “(x+2)(x+3)-(x-2)(x-5)=-4 (x+1)( $x^{2}$ -x+1)-x(x+3)(x-3)=8”

1. Đáp án:

    

   Giải thích các bước giải:

    $a)(x+2).(x+3)-(x-2).(x-5)=-4$

   $⇔x^2+5x+6-(x^2-7x+10)=-4$

   $⇔x^2+5x+6-x^2+7x-10=-4$

   $⇔12x-4=-4$

   $⇔12x=0$

   $⇔x=0$

   $b)(x+1).(x^2-x+1)-x.(x+3).(x-3)=8$

   $⇔x^3+1-x.(x^2-9)=8$

   $⇔x^3+1-x^3+9x=8$

   $⇔9x=7$

   $⇔x=\dfrac{7}{9}$

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2. `(x+2)(x+3)-(x-2)(x-5)=-4`

   ⇔` (x²+3x+2x+6)` `-“( x²-5x-2x+10) “=-4`

   `⇔` `x² +5x +6 – x²+7x -10 +4=0`

   `⇔` `12x=0`

   `⇔ ` `x=0`

    Vậy `x=0`

   `(x+1)( x² -x+1)-x(x+3)(x-3)=8`

   `⇔` `(x³+1)³` `-x(x²-9) =8`

   `⇔` `x³ +1 -x³ +9x-8=0`

   `⇔` `9x -7=0`

   `⇔` `9x=7`

   `⇔“ x=“7/9`

     Vậy `x=“7/9`

   @lien2k6

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