2(x-8)+x=7 4x(x-3)-5(3-x)=0 (2x+1)²-(4x-1)(x+2)=0

2 bình luận về “2(x-8)+x=7 4x(x-3)-5(3-x)=0 (2x+1)²-(4x-1)(x+2)=0”

1. 2 ( x – 8 ) + x = 7

   ⇔ 2x – 16 + x = 7

   ⇔ 3x = 7 + 16

   ⇔ 3x = 23

   ⇔ x = 23/3

   Vậy x = 23/3

   _________

   4x ( x – 3 ) – 5 ( 3 – x ) = 0

   ⇔ 4x ( x – 3 ) + 5 ( x – 3 ) = 0

   ⇔ ( x – 3 ) ( 4x + 5 ) = 0

   TH1:

   x –  3 = 0

   ⇔ x = 3

   TH2:

   4x + 5 = 0

   ⇔ 4x = -5

   ⇔ x = -5/4

   Vậy x ∈ { 3 ; -5/4 }

   ______

   ( 2x + 1 )^2 – ( 4x – 1 ) ( x + 2 ) = 0

   ⇔ ( 2x )^2 + 2 . 2x . 1 + 1^2 – ( 4x^2 + 8x – x – 2 ) = 0

   ⇔ 4x^2 + 4x + 1 – 4x^2 – 8x + x + 2 = 0

   ⇔ -3x = -3

   ⇔ x = 1

   Vậy x = 1

   Trả lời
2. 2(x-8)+x=7

   =>2x+2.(-8)+x=7

   =>2x-16+x=7

   =>3x=7+16

   =>3x=232

   =>x=23/3

   Vậy x=23/3

   $\\$

   4x(x-3)-5(3-x)=0

   =>4x^2-7x-15=0

   =>(x-4)(4x+5)=0

   => \(\left[ \begin{array}{l}x-3=0\\3x+5=0\end{array} \right.\) 

   => \(\left[ \begin{array}{l}x=3\\x=-\dfrac{5}{4}\end{array} \right.\) 

   Vậy x=3;-5/4

   $\\$

   (2x+1)^2-(4x-1)(x+2)=0

   =>4x^2+4x+1+(-4x)x+(-4x).2+1x+1.2=0

   =>3x=3

   =>x=1

   Vậy x=1

   Trả lời