\(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)

=> \(2x^2+3\left(x^2-1\right)=5x^2+5x\)

=> \(2x^2+3x^2-3-5x^2-5x=0\)

=> \(-3-5x=0\)

=> \(5x=-3\Rightarrow x=-\frac{3}{5}\)

\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+\frac{9}{2}\right)=\frac{7}{2}\)

=> \(x\left[2x\left(x+5\right)-1\left(x+5\right)\right]-2x^2\left(x+\frac{9}{2}\right)-1\left(x+\frac{9}{2}\right)=\frac{7}{2}\)

=> \(x\left(2x^2+10x-x-5\right)-2x^3-9x^2-x-\frac{9}{2}=\frac{7}{2}\)

=> \(2x^3+10x^2-x^2-5x-2x^3-9x^2-x-\frac{9}{2}=\frac{7}{2}\)

=> \(\left(2x^3-2x^3\right)+\left(10x^2-x^2-9x^2\right)+\left(-5x-x\right)-\frac{9}{2}=\frac{7}{2}\)

=> \(-6x-\frac{9}{2}=\frac{7}{2}\)

=> \(-6x=8\Rightarrow x=-\frac{8}{6}=-\frac{4}{3}\)

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

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

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

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

=> 83x + 5 - 7 = 81

=> 83x = 81 + 7 - 5

=> 83x = 83

=> x = 1

1) \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)

\(\Leftrightarrow2x^2+3x^2-3-5x^2-5x=0\)

\(\Leftrightarrow5x=-3\)

\(\Rightarrow x=-\frac{3}{5}\)

2) \(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+\frac{9}{2}\right)=\frac{7}{2}\)

\(\Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-\frac{9}{2}=\frac{7}{2}\)

\(\Leftrightarrow-6x=8\)

\(\Rightarrow x=-\frac{4}{3}\)

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

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

\(\Leftrightarrow83x=83\)

\(\Rightarrow x=1\)

a) 2x² + 3( x - 1 )( x + 1 ) = 5x( x + 1 )

<=> 2x² + 3( x² - 1 ) = 5x² + 5x

<=> 2x² + 3x² - 5x² - 5x = 3

<=> -5x = 3

<=> x = -3/5

b) x( 2x - 1 )( x + 5 ) - ( 2x² + 1 )( x + 9/2 ) = 7/2

<=> x( 2x² + 9x - 5 ) - ( 2x³ + 9x² + x + 9/2 ) = 7/2

<=> 2x³ + 9x² - 5x - 2x³ - 9x² - x - 9/2 = 7/2

<=> -6x - 9/2 = 7/2

<=> -6x = 8

<=> x = -8/6 = -4/3

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

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

<=> 83x - 2 = 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

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

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

\(\Leftrightarrow83x=83\)

\(\Leftrightarrow x=1\)

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

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

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

\(83x-2=81\)

\(83x=83\)

\(x=1\)

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

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

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

1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)

\(\Leftrightarrow6x=6\)

hay x=1