\(a,3.\left(2x-3\right)+4.\left(3x-3\right)=19\)

\(\Leftrightarrow6x-9+12x-12=19\)

\(\Leftrightarrow18x-21=19\)

\(\Leftrightarrow18x=40\)

\(\Leftrightarrow x=\frac{20}{9}\)

\(b,4\cdot\left(2x-1\right)-3\cdot\left(2x+3\right)=13\)

\(\Leftrightarrow8x-4-6x-9=13\)

\(\Leftrightarrow2x-13=13\)

\(\Leftrightarrow2x=26\)

\(\Leftrightarrow x=13\)

\(c,5\cdot\left(6x-3\right)-3\cdot\left(4x-5\right)=19\)

\(\Leftrightarrow30x-15-12x+15=19\)

\(\Leftrightarrow18x=19\)

\(\Leftrightarrow x=\frac{19}{18}\)

Kết bạn với mình nha

A) Ta có :  3(2x - 3) + 4(3x - 3) = 19

<=> 6x - 9 + 12x - 12 = 19

<=> 18x - 21 = 19

=> 18x = 40

=> x = \(\frac{20}{9}\)

\(\dfrac{2x+19}{21}-\dfrac{2x+17}{23}=\dfrac{2x+7}{33}-\dfrac{2x+5}{35}\)

\(\Rightarrow\dfrac{2x+19}{21}-\dfrac{2x+17}{23}-\dfrac{2x+7}{33}+\dfrac{2x+5}{35}=0\)

\(\Rightarrow\left(\dfrac{2x+19}{21}+1\right)-\left(\dfrac{2x+17}{23}+1\right)-\left(\dfrac{2x+7}{33}+1\right)+\left(\dfrac{2x+5}{35}+1\right)=0\)

\(\Rightarrow\dfrac{2x+40}{21}-\dfrac{2x+40}{23}-\dfrac{2x+40}{33}+\dfrac{2x+40}{35}=0\)

\(\Rightarrow\left(2x+40\right)\left(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}\right)=0\)

\(\Rightarrow2x+40=0\Rightarrow x=-20\)( do \(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}>0\))

Ta có: \(\dfrac{2x+19}{21}-\dfrac{2x+17}{23}=\dfrac{2x+7}{33}-\dfrac{2x+5}{35}\)

\(\Leftrightarrow\left(2x+40\right)\left(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}\right)=0\)

\(\Leftrightarrow2x+40=0\)

hay x=-20

\(\dfrac{x+1}{1}+\dfrac{2x+3}{3}+\dfrac{3x+5}{5}+...+\dfrac{10x+19}{19}=12+\dfrac{4}{3}+\dfrac{6}{5}+...+\dfrac{20}{19}\)

\(x+1+\dfrac{2x}{3}+1+\dfrac{3x}{5}+1+...+\dfrac{10x}{19}+1-12-\dfrac{4}{3}-\dfrac{6}{5}-...-\dfrac{20}{19}=0\)

\(x+\dfrac{2x}{3}-\dfrac{4}{3}+\dfrac{3x}{5}-\dfrac{6}{5}+...+\dfrac{10x}{19}-\dfrac{20}{19}+10-12=0\)

\(x-2+\dfrac{2x-4}{3}+\dfrac{3x-6}{5}+...+\dfrac{10x-20}{19}=0\)

\(x-2+\dfrac{2\left(x-2\right)}{3}+\dfrac{3\left(x-2\right)}{5}+...+\dfrac{10\left(x-2\right)}{19}=0\)

\(\left(x-2\right)\left(\dfrac{2}{3}+\dfrac{3}{5}+...+\dfrac{10}{19}\right)=0\)

Ta thấy \(\left(\dfrac{2}{3}+\dfrac{3}{5}+...+\dfrac{10}{19}\right)>0\)

\(\Rightarrow x-2=0\Rightarrow x=2\)

=>6x-19=3x+5

=>3x=24

hay x=8

Sửa đề: 2(x-1)^2+4x-19=(2x-1)(2x+5)

=>2(x^2-2x+1)+4x-19=4x^2+10x-2x-5

=>2x^2-4x+2+4x-19=4x^2+8x-5

=>4x^2+8x-5=2x^2-17

=>2x^2+8x+12=0

=>x^2+4x+6=0

=>(x+2)^2+2=0(loại)

1, 

2x + 5 = 19

2x       = 19 - 5

2x       = 14

  x       = 14 : 2

  x        = 7

2,

( 1 - 2x ) + 5 = 10

( 1 - 2x )       = 5

       2x         = 1 - 5

       2x         = -4

         x         = -4 : 2

         x         = -2

1)    2x + 5 = 19

2x = 19 - 5

2x = 14 

x = 14 : 2

x= 7

2)    (1-2x)  + 5 = 10

(1- 2x) = 10 - 5

(1-2x) = 5

2x = 1-5

2x = -4

x= -4 : 2

x = -2

1) 2x +5 =19                  

   2x      =19-5

  2x       =14

   x       =14:2

  x       =7

2) (1-2x)+5=10

     (1-2x)   =10-5

    (1-2x)   =5

       2x     =1-5

       2x    =-4

        x    =(-4):2

      x      = -2

k mik nha

Đáp án C là đúng nha bạn 😉

Dòng (C) nhé. Phân tích ra và xét theo các số lẻ.