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

\(\Rightarrow\left(48x^2-12x-20x+5\right)\left(3x-7\right)\left(1-16x\right)=81\)

\(\Rightarrow\left(48x^2-32x+5\right)\left(3x-7\right)\left(1-16x\right)=81\)

\(\Rightarrow\left(144x^3-336x^2-96x^2+224x+15x-35\right)\left(1-16x\right)=81\)

\(\Rightarrow\left(144x^3-432x^2+239x-35\right)\left(1-16x\right)=81\)

\(\Rightarrow144x^3-2304x^4-432x^2+6912x^3+239x-3824x^2-35+560x=81\)

\(\Rightarrow-2304x^4+7056x^3-4256x^2+799x-116=0\)

\(\Rightarrow\left[{}\begin{matrix}x_1\approx2,3\\x_2\approx0,5\end{matrix}\right.\)

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

48x² - 32x + 5 - 48x + 115x - 7     = 81

83x - 2                                        = 81

83x                                             = 83

=> x                                           = 1

Sửa lại dòng 2 : 48x² - 32x + 5 - 48x² + 115x - 7 = 81

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

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

  => 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

83x = 81 - 5 + 7

83x = 83

x = 83/83

x = 1

Chúc bn hok tốt nha

1) Ta có: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(5-x\right)^2\)

\(\Leftrightarrow x^2+2x+5x+10-12x+9=25-10x+x^2\)

\(\Leftrightarrow x^2-5x+19-25+10x-x^2=0\)

\(\Leftrightarrow5x-6=0\)

\(\Leftrightarrow5x=6\)

\(\Leftrightarrow x=\frac{6}{5}\)

Vậy: \(x=\frac{6}{5}\)

2) Ta có: \(\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-8\)

\(\Leftrightarrow x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)=12x^2-12x-8\)

\(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2+12x+8=0\)

\(\Leftrightarrow12x+24=0\)

\(\Leftrightarrow12x=-24\)

\(\Leftrightarrow x=-2\)

Vậy: x=-2

3) Ta có: \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)

\(\Leftrightarrow36x^2-12x-36x^2+27x-30=0\)

\(\Leftrightarrow15x-30=0\)

\(\Leftrightarrow15x=30\)

\(\Leftrightarrow x=2\)

Vậy: x=2

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

\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x-81=0\)

\(\Leftrightarrow83x-83=0\)

\(\Leftrightarrow83x=83\)

\(\Leftrightarrow x=1\)

Vậy: 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

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

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

\(\Leftrightarrow83x=83\)

\(\Leftrightarrow x=1\)

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

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

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

\(83x-2=81\)

\(83x=83\)

\(x=1\)