(3x+2).(x+1)=3x.(5+x)                                                                               

\(\Rightarrow\)\(3x^2+3x+2x+2=15x+3x^2\)                               

\(\Rightarrow3x^2+5x+2=15x+3x^2\)

\(\Rightarrow5x-15x+2=3x^2-3x^2\)

\(\Rightarrow-10x+2=0\)

\(-10x=-2\)

    \(x=\frac{1}{5}\)

3x(x-5)-2(x-6)=x(5+3x)

(3x²-15x)-(2x-12)=5x+3x²

3x²-15x-2x+12=5x+3x²

3x²-15x-2x-5x-3x²=12

-22x=12

x=6/-11

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

\(\Leftrightarrow\left(9x^2-2^2\right)-\left(9x^2-6x+1\right)=5\)

\(\Leftrightarrow9x^2-4-9x^2+6x-1-5=0\)
\(\Leftrightarrow6x=10\)

\(\Leftrightarrow x=\dfrac{5}{3}\)

Vậy \(S=\left\{\dfrac{5}{3}\right\}\)

(3x + 2) . (3x- 2)- (3x- 1)^2= 5

<=> (3x + 2) . (3x- 2)- [ ( 3x^2 )  - 2 . 3x .1 + 1^2  ] = 5

<=>   9x^2 - 6x + 6x - 4 - ( 9x^2 - 6x + 1 ) = 5

<=>    9x^2 - 6x + 6x - 4 - 9x^2 + 6x - 1 = 5

<=>  6x - 5 = 5

<=> 6x = 5 + 5

<=> 6x = 10

<=> x = 10/6

<=> x = 5/3 

Đặt `3(x+2)-1/3(6-3x)=0`

`<=>3(x+2)-(2-x)=0`

`<=>3x+2+x-2=0`

`<=>4x=0`

`<=>x=0`

Vậy nghiệm của đa thức là 0

`3x(x-5)-(x+3x)=0`

`<=>3x(x-5)-4x=0`

`<=>x(3x-15-4)=0`

`<=>x(3x-19)=0`

`<=>[(x=0),(3x-19=0):}`

`<=>[(x=0),(x=19/3):}`

Vậy nghiệm đa thức là 0 và `19/3`.

a) Đặt \(3\left(x+2\right)-\dfrac{1}{3}\left(6-3x\right)=0\)

\(\Leftrightarrow3x+6-2+x=0\)

\(\Leftrightarrow4x=-4\)

hay x=-1

b) Đặt 3x(x-5)-(x+3x)=0

\(\Leftrightarrow3x^2-15x-4x=0\)

\(\Leftrightarrow x\left(3x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{3}\end{matrix}\right.\)

a: \(\Leftrightarrow-x^2-3x+x+3+x^2-6x=11\)

=>-8x+3=11

=>-8x=8

hay x=-1

b: \(\Leftrightarrow3x^2-15x+x-5-3x^2+3x=5\)

=>-11x=10

hay x=-10/11

\(\Leftrightarrow3x^2+3x-3x^2-2x=5\)

=>3x-2x=5

hay x=5

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

\(x=5\)

\(\left(5-x\right)\left(x-2\right)+\left(x-7\right)\left(x+7\right)=\left(3x-1\right)^2-\left(3x-2\right)\left(3x+2\right)\\ \Leftrightarrow-x^2+7x-10+x^2-49=9x^2-6x+1-9x^2+4\\\Leftrightarrow7x-59=-6x+5\\ \Leftrightarrow13x=44\\ \Leftrightarrow x=\dfrac{64}{13} \)

 sai đề mình chữa lại cho (3x-5)²-x(3x-5)=0

  (3x-5)^2-x(3x-5)=0

  3x²-25-3x²-5x=0

  (3x²-3x²)-25-5x=0

             -25-5x=0

             5x=-25-0

             5x=-25

               x=-25:5=-5

a) \(\Leftrightarrow4x^2-10x-4x^2-3x=19\\ \Leftrightarrow-13x=19\\ \Leftrightarrow x=-\dfrac{19}{13}\)

b) \(\Leftrightarrow\left(x-7\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{5}{3}\end{matrix}\right.\)