`x - ( 2x - 1 ) <= 3x - 3`

`<=> x - 2x + 1 <= 3x - 3`

`<=> 3x - x + 2x >= 1 + 3`

`<=> 4x >= 4`

`<=> x >= 1`

Vậy `S = { x | x >= 1 }`

\(\Leftrightarrow x-2x+1\le3x-3\)

\(\Leftrightarrow-4x\le-4\)

\(\Leftrightarrow x\ge1\)

`x - (2x - 1) <= 3x - 3`

`<=> x - 2x + 1 <= 3x - 3`

`<=> 3x - x + 2x >= 1 + 3`

`<=> 4x >= 4`

`<=>` `x >= 1`

Vậy `S = {x | x >= 1}`

x(x²+6x+9) - 3x= x³+6x²+12x+8+1

\(\Leftrightarrow\)x³+6x²+9x-3x=x³+6x²+12x+9

\(\Leftrightarrow\)6x=12x+9

\(\Leftrightarrow\)6x=-9

\(\Leftrightarrow\)x=-3/2

Vậy phương trình có 1 nghiệm duy nhất x=-3/2

x(x + 3)^2  - 3x = (x + 2)^3 + 1

<=> x(x^2 + 6x + 9) = x^3 + 6x^2 + 12x + 8 + 1

<=> x^3 + 6x^2 + 9x = x^3 + 6x^2 + 12x + 9

<=> 3x + 9 = 0

<=> 3x = -9

<=> x = -3

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

\(\Leftrightarrow x\left(x^2+6x+9\right)=x^3+6x^2+12x+8+1\)

\(\Leftrightarrow x^3+6x^2+9x=x^3+6x^2+12x+9\)

\(\Leftrightarrow3x+9=0\)

\(\Leftrightarrow3x=-9\)

\(\Leftrightarrow x=-3\)

`a)(2x-1)^2-0,25=0`

`<=>(2x-1-0,5)(2x-1+0,5)=0`

`<=>(2x-1,5)(2x-0,5)=0`

`<=>[(x=0,75)(x=0,25):}`

`b)x^2+9=6x`

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

`<=>x-3=0`

`<=>x=3`

`c)(x^2-4)-3x-6=0`

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

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

`<=>(x+2)(x-5)=0`

`<=>[(x=-2),(x=5):}`

a: =>(2x-1-0,5)(2x-1+0,5)=0

=>(2x-1,5)(2x-0,5)=0

=>x=0,25 hoặc x=0,75

b: =>x^2-6x+9=0

=>(x-3)^2=0

=>x-3=0

=>x=3

c: =>(x-2)(x+2)-3(x+2)=0

=>(x+2)(x-5)=0

=>x=5 hoặc x=-2

`x(x - 4) - 3x + 12 = 0`

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

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

`<=>` $\left[\begin{matrix} x + 3 = 0\\ x - 4 = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = -3\\ x = 4\end{matrix}\right.$

Vậy `S = {-3; 4}`

\(x\left(x-4\right)-3x+12=0\)

\(\Leftrightarrow x^2-4x-3x+12=0\)

\(\Leftrightarrow x\left(x-4\right)-3\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)

b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)

c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)

d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)

a) Ta có: 4x-20=0

$\iff 4x=20$

hay x=5

Vậy: S={5}

b) Ta có: $2x+x+12=0$

$\iff 3x+12=0$

$\iff 3x=-12$

hay x=-4

ăn lồn đê

Đúng làm trẻ trâu , ăn nói mất lịch sự

\(\left(2x-3\right)\cdot4,8=\left(3x+1\right)\cdot\left(-2,4\right)\)

\(9,6x-14,4=-7,2x-2,4\)

\(9,6x+7,2x=14,4-2,4\)

\(16,8x=12\)

\(x=\dfrac{12}{16,8}=\dfrac{5}{7}\)

x = 5/7