$ a/ 12x(x – 5) – 3x(4x - 10) = 120$

`<=>12x^2-60x-12x^2+30x=120`

`<=>-30x=120`

`<=>x=-4`

Vậy `x=-4`

$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$

`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`

`<=>-6x^2+26x=112-6x^2-2x`

`<=>28x=112`

`<=>x=4`

Vậy `x=4`

$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$

`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`

`<=>-32x-18x^2=154+45x-18x^2`

`<=>77x=-154`

`<=>x=-2`

Vậy `x=-2`

giải chi tiết ra giúp mk nhé, cảm ơn nhiều

a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0

=>-5x-4=0

=>x=-4/5

b: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81

=>83x=83

=>x=1

a,3x-0,75=-1,2

3x=-1,2+-0,75

3x=-9/20

x=-9/20:3

x=-3/20

kl:....

b,2/3/5:(2x-1/2)=1,3

2/3/5??????

c,3/4x-5/6=7/12x-5/4

3/4x-7/12x=-5/4+5/6

9/12x-7/12x=-15/12+10/12

2/12x=-5/12

2:12x=-5/12

12x=2:-5/12

12x=-24/5

x=-24/5:12

x=-2/5

kl:....

chúc bn học tốt

a: \(\Leftrightarrow3x+7\in\left\{1;-1;3;-3;11;-11;33;-33\right\}\)

hay \(x=-6\)

b: \(\Leftrightarrow3x-1\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{0;-2\right\}\)

\(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{\left(x-2\right)\left(3x+2\right)}+\dfrac{1}{\left(x-2\right)\left(x-10\right)}\right)=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x-2\right)}\left(\dfrac{x-10+3x+2}{\left(3x+2\right)\left(x-10\right)}\right)\)

\(=\lim\limits_{x\rightarrow2}\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(3x+2\right)\left(x-10\right)}=\lim\limits_{x\rightarrow2}\dfrac{4}{\left(3x+2\right)\left(x-10\right)}=-\dfrac{1}{16}\)

a)\(60x^2+35x-60x^2+15x=100\)

  35x+15x=100

  50x=100   =>x=2

b)\(10x^2-35x+16x-10x^2=5\)

  -35x+16x=5

   -19x=5    =>x=-5/19

a)

  35x+15x=100

  50x=100   

=>x=2

b)

  -35x+16x=5

   -19x=5   

=>x=-5/19