2(x-5) – x ( x+ 1) = x ( 5-x) b,x(x-5)-(x+2) (x-3(=6 c, (2x+1)(x-3)=2x (x+5)-4x d, (x-5)(x+1) – ( x-2)(x+3)

0 bình luận về “2(x-5) – x ( x+ 1) = x ( 5-x) b,x(x-5)-(x+2) (x-3(=6 c, (2x+1)(x-3)=2x (x+5)-4x d, (x-5)(x+1) – ( x-2)(x+3)”

1. a.2(x-5) – x ( x+ 1) = x ( 5-x)
   <=> 2x – 10 – x^2 – x = 5x – x^2
   <=> x – x^2 – 5x  + x^2 = 10
   <=> -4x = 10
   <=> x = -5/2
   b,x(x-5)-(x+2) (x-3(=6
   <=> x^2 – 5x – x^2 + x + 6 = 6
   <=> -4x = 0
   <=> x = 0

   c, (2x+1)(x-3)=2x (x+5)-4x
   <=> 2x^2 – 5x – 3 = 2x^2 + 10x – 4x
   <=> 2x^2 – 5x – 2x^2 – 6x = 3
   <=> -11x = 3
   <=> x = -3/11
   d, (x-5)(x+1) – ( x-2)(x+3)
   = x^2 – 4x – 5 – x^2 – x + 6
   = -5x + 1

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2. $a)$

   $2(x-5) – x ( x+ 1) = x ( 5-x)$

   $⇔2x-10-x^{2}-x=5x-x^{2}$

   $⇔2x-x-5x-x^{2}+x^{2}=10$

   $⇔-4x^{}=10$

   $⇔x=\frac{-5}{2}$

         $Vậy$ $x=\frac{-5}{2}$

   $b)$

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

   $⇔x^{2}-5x-x^{2}+3x-2x+6=6$

   $⇔-4x^{}=0$

   $x=0$

         $Vậy$ $x=0$

   $c)$

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

   $⇔2x^{2}-6x+x-3=2x^{2}+10x-4x$

   $⇔2x^{2}-6x+x-2x^{2}-10x+4x=3$

   $⇔-11x=3$

   $⇔x= \frac{-3}{11} $

          $Vậy$ $x= \frac{-3}{11} $

   $d)$

   $(x-5)(x+1) – ( x-2)(x+3)$

   $=x^{2}+x-5x-5-x^{2}+2x-3x+6$ 

   $=-5x+1^{}$

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