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

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

1: Sửa đề: 3x-5

\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)

2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

=5x^2+14x^2+12x+8

3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)

4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)

5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)

Bài 3: 

\(\Leftrightarrow x^3+64-x^3+25x=264\)

hay x=8

\(1,C=6x^2+23x-55-6x^2-23x-21=-76\\ 2,=\left(2x^4-x^2+2x^3-x-6x^2+6-3\right):\left(2x^2-1\right)\\ =\left[\left(2x^2-1\right)\left(x^2+x-6\right)-3\right]:\left(2x^2-1\right)\\ =x^2+x-6\left(dư.-3\right)\\ 3,\Leftrightarrow x^3+64-x^3+25x=264\\ \Leftrightarrow25x=200\Leftrightarrow x=8\)

a: Ta có: \(2x^3-18x=0\)

\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)

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

b: Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)

\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)

\(\Leftrightarrow-13x=13\)

hay x=-1

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

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8=3-3x^2\)

\(\Leftrightarrow3x=12\)

hay x=4

a) 2x³-18x=0

⇔ 2x(x²-9)=0

⇔ 2x(x-3)(x+3)=0

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

b)(3x-1)(2x+1)-6x(x+2)=11

⇔ 6x²+x-1-6x²-12x=11

⇔ -11x=12

\(\Leftrightarrow x=-\dfrac{12}{11}\)

c) (x-1)³-(x+2).(x²-2x+4)=3.(1-x²)

⇔ x³-3x²+3x-1-x³-8-3+3x²=0

⇔ 3x=12

⇔   x=4

c. (x - 1)³ - (x + 2)(x² - 2x + 4) = 3(1 - x²)

<=> (x³ - 3x² + 3x - 1) - (x³ - 2x² + 4x + 2x² - 4x + 8) = 3 - 3x²

<=> x³ - 3x² + 3x - 1 - x³ + 2x² - 4x - 2x² + 4x - 8 = 3 - 3x²

<=> x³ - x³ - 3x² + 2x² - 2x² + 3x² + 3x - 4x + 4x = 3 + 1 + 8

<=> 3x = 12

<=> x = 4

`@` `\text {Ans}`

`\downarrow`

`a)`

`3x ( 12x - 4 ) - 9x( 4x - 3 ) = 30`

`=> 3x (12x-4) - 3*3x (4x - 3) = 30`

`=> 3x [12x - 4 - 3(4x-3)] = 30`

`=> 3x (12x - 4 - 12x + 9) = 30`

`=> 3x (-4+9)=30`

`=> 3x*5=30`

`=> 3x=6`

`=> x=2`

Vậy, `x=2`

`b)`

`x( 5 - 2x) + 2x( x - 1)`

`=> x(5-2x) + 2x^2 - 2x=15`

`=> 5x - 2x^2 + 2x^2 - 2x =15`

`=> 3x = 15`

`=> x=5`

Vậy, `x=5.`

a: =>36x^2-12x-36x^2+27x=30

=>15x=30

=>x=2

b: =>5x-2x^2+2x^2-2x=15

=>3x=15

=>x=5

c) Ta có: \(\dfrac{5x^4+9x^3-2x^2-4x-8}{x-1}\)

\(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

\(=\dfrac{5x^3\left(x-1\right)+14x^2\left(x-1\right)+12x\left(x-1\right)+8\left(x-1\right)}{x-1}\)

\(=5x^3+14x^2+12x+8\)

d) Ta có: \(\dfrac{5x^3+14x^2+12x+8}{x+2}\)

\(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}\)

\(=\dfrac{5x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)}{x+2}\)

\(=5x^2+4x+4\)

c) Ta có: \(\dfrac{5x^4+9x^3-2x^2-4x-8}{x-1}\)

\(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

\(=\dfrac{5x^3\left(x-1\right)+14x^2\left(x-1\right)+12x\left(x-1\right)+8\left(x-1\right)}{x-1}\)

\(=5x^3+14x^2+12x+8\)

b: \(\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)

\(=x^2-2x+1\)

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

c: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

\(=5x^3+14x^2+12x+8\)

\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)

a) 9x²-49=0

(3x)²-7²=0

<=> (3x-7)(3x+7)=0

th1: 3x-7=0

<=>3x=7

<=>x=\(\dfrac{7}{3}\)

th2: 3x+7=0

<=>3x=-7

<=>x=\(-\dfrac{7}{3}\)

a: Ta có: \(x\left(2-x\right)+\left(x^2+x\right)=7\)

\(\Leftrightarrow2x-x^2+x^2+x=7\)

\(\Leftrightarrow3x=7\)

hay \(x=\dfrac{7}{3}\)

b: Ta có: \(\left(2x+1\right)^2-x\left(4-5x\right)=17\)

\(\Leftrightarrow4x^2+4x+1-4x+5x^2=17\)

\(\Leftrightarrow9x^2=16\)

\(\Leftrightarrow x^2=\dfrac{16}{9}\)

hay \(x\in\left\{\dfrac{4}{3};-\dfrac{4}{3}\right\}\)

c: Ta có: \(\left(x-4\right)^2-\left(2x+1\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

d: ta có: \(\dfrac{2x^3-8x^2+10x}{2x}=0\)

\(\Leftrightarrow x^2-4x+5=0\)

\(\Leftrightarrow\left(x-2\right)^2+1=0\)(vô lý)