Giải phuơng trình sau a) 4x-11=8x-24 b) 15-3x=2x+ 20 c) (x+6) ² – ( x-2)(X+2)=0 d) (4x-5)(4x+1)-(4x-3)^2= 0 e) (2x-5)(3x+1)-6(x-2)(x+2) = 0 f) (x-3)(

0 bình luận về “Giải phuơng trình sau a) 4x-11=8x-24 b) 15-3x=2x+ 20 c) (x+6) ² – ( x-2)(X+2)=0 d) (4x-5)(4x+1)-(4x-3)^2= 0 e) (2x-5)(3x+1)-6(x-2)(x+2) = 0 f) (x-3)(”

1. a) <=> 4x-8x= -24+11

   <=> -4x = -13

   <=> x = (-13): (-4 )

   vậy S = {13 phần 4}

   b) <=> -3x-2x = 20+ (-15)

   <=>  -5x = 5

   <=> x= 5: (-5)

   vậy S= {-1}

   c) <=>  (x+6)^2 – (x+2)^2=0

   <=>  x+6 =0    => x= -6

   <=>  x-2 =0     => x= 2

   vậy S=  { -6;2}

   f)  <=>(x-3)^2 – (x-1)^2 =0

   <=>  x-3  =0 => x=3

   <=>  x+1 =0 => x=-1

   vậy S = {3;-1}
    Còn mấy câu nx ko bt làm 

   sorry 🙂

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2. a) 4x-11=8x-24

   ⇔ 4x-8x=-24+11

   ⇔ -4x=-13

   ⇔ x=$\frac{13}{4}$  

   b) 15-3x=2x+ 20

   ⇔ -3x-2x=20-15

   ⇔ -5x=5

   ⇔ x=-1

   c) (x+6) ² – ( x-2)(X+2)=0

   ⇔ x² + 12x + 36 – x² + 4=0

   ⇔ 12x+40=0

   ⇔ 12x=40

   ⇔ x=$\frac{40}{12}$ =$\frac{10}{3}$ 

   d) (4x-5)(4x+1)-(4x-3)^2= 0

   ⇔ 16x² + 4x – 20x – 5 – 16x² + 24x -9=0

   ⇔ 8x – 14 = 0

   ⇔ 8x = 14

   ⇔ x = $\frac{7}{4}$ 

   e) (2x-5)(3x+1)-6(x-2)(x+2) = 0

   ⇔ 6x² + 2x – 15x – 5 – 6x² + 24 = 0

   ⇔ -13x +19 = 0

   ⇔ -13x = -19

   ⇔ x = $\frac{13}{19}$ 

   f) (x-3)(x+3)- (x-1) ²=0

   ⇔ x² – 9 – x² + 2x – 1 = 0

   ⇔ 2x – 10 = 0

   ⇔ 2x = 10

   ⇔ x = 5

   g) 4x – $\frac{1}{9}$ – $\frac{2x}{3}$ = $\frac{1}{6}$ – $\frac{3x}{2}$

   ⇔ 4x – $\frac{2x}{3}$ + $\frac{3x}{2}$ = $\frac{1}{6}$ + $\frac{1}{9}$

   ⇔ $\frac{24x-4x+9x}{6}$ = $\frac{3+2}{18}$

   ⇔ $\frac{29x}{6}$ = $\frac{5}{18}$

   ⇔ $\frac{87x}{18}$ = $\frac{5}{18}$

   ⇔ 87x = 5

   ⇔ x = $\frac{5}{87}$ 

   h)4 – $\frac{3x}{6}$ – 2x = $\frac{3}{4}$ – 1 + $\frac{5x}{3}$

   ⇔ $\frac{-3x}{6}$ – 2x – $\frac{5x}{3}$ = $\frac{3}{4}$ – 1 – 4

   ⇔ $\frac{-3x – 12x -10x}{6}$ = $\frac{3 – 4 – 16}{4}$

   ⇔ $\frac{-25x}{6}$ = $\frac{-17}{4}$

   ⇔ $\frac{-50x}{12}$ = $\frac{-51}{12}$4x

   ⇔ 50x = 51

   ⇔ x = 51/50

   i) 2/3.( 1-2x) = 3/4-5/6.( 2x+3)

   ⇔ 2/3 – 4x/3 = 3/4 – 5x/3 – 5/2

   ⇔ -4x/3 + 5x/3 = 3/4 – 5/2 – 2/3

   ⇔ x/3 = -29/12

   ⇔ 4x/12 = -29/12

   ⇔ 4x = -29

   ⇔ x = -29/4

   i) 4/9(2x+3) – 1/2 = x – 5/6(4x-1)

   ⇔ 8x/9 + 4/3 – 1/2 = x – 10x/3 + 5/6

   ⇔ 8x/9 + 10x/3 – x = 5/6 -4/3 + 1/2

   ⇔ 8x+30x-9x/9 = 0

   ⇔ 29x = 0

   ⇔ x=0

   Xin ctlhn ạ!

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