(12x - 5) (4x - 1) + (3x - 7) (1 - 16x) = 81

=> 48x² - 12x - 20x + 5 + 3x - 48x² - 7 + 112x = 81

=> 83x = 83

=> x = 1

\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)

\(\text{⇔}48x^2-32x+5-48x^2-7+115x=81\)

\(\text{⇔}83x-2=81\)

\(\text{⇔}83x=83\)

\(\text{⇔}x=1\)

Vậy: x=1

Ta có: \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)

\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x=81\)

\(\Leftrightarrow83x=83\)

hay x=1

\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)

\(\Leftrightarrow48x^2-32x+5+48x^2+115x-7=81\)

\(\Leftrightarrow83x=83\)

\(\Leftrightarrow x=1\)

\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x=81\\ \Leftrightarrow83x=83\Leftrightarrow x=1\)

Rút gọn vế trái:

VT = (12x – 5)(4x – 1) + (3x – 7)(1 – 16x)

     = 12x.(4x – 1) + (–5).(4x – 1) + 3x.(1 – 16x) + (–7).(1 – 16x)

     = 12x.4x+ 12x.(–1) + (–5).4x + (–5).(–1) + 3x.1 + 3x.(–16x) + (–7).1 + (–7).(–16x)

     = 48x2 – 12x – 20x + 5 + 3x – 48x2 – 7 + 112x

     = (48x2 – 48x2) + (– 12x – 20x + 3x + 112x) + (5 – 7)

     = 83x – 2

Vậy ta có:

83x – 2 = 81

       83x = 81 + 2

       83x = 83

           x = 83 : 83

           x = 1.

(12x - 5)(4x - 1) + (3x - 7)(1 - 16x) = 81

<=> 48x² - 12x - 20x + 5 + 3x - 48x² - 7 + 112x = 81

<=> 48x² - 48x² - 12x - 20x + 3x + 112x = 81 - 5 + 7

<=> 83x = 83

<=> x = 1

\(\Rightarrow48x^2-32x+5-48x^2+115x-7=81\)

\(\Rightarrow83x=83\Rightarrow x=1\)

ê m nãy t tính ra \(\dfrac{2}{83}\) đến lúc thử lại kq thì ......

a)\(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)

    \(4x^2-20x-\left(4x^2-7x+3\right)=5\)

    \(4x^2-20x-4x^2+7x-3=5\)

     \(-13x=8\)

      \(x=-\frac{8}{13}\)

b)\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)

     \(48x^2-32x+5+3x-48x^2-7+112x=81\)

     \(83x-2=81\)

     \(x=1\)

Ta có : ( 12x - 5 ) ( 4x - 1 ) + ( 3x - 7 ) ( 1 -16x ) = 81

=> 48x² - 20x - 12x + 5 + 3x - 7 - 48x² + 112x = 81

=> 80x - 2 = 81

=> 80x = 83

=> x = 83/80