a) \(\left(3x+5\right)\left(7-2x\right)+6x\left(x+4\right)=0\)

\(\Leftrightarrow21x-6x^2+35-10x+6x^2+24x=0\)

\(\Leftrightarrow35x=-35\Leftrightarrow x=-1\)

b) \(x^3-25x=0\)

\(\Leftrightarrow x\left(x^2-25\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

a: Ta có: \(\left(3x+5\right)\left(7-2x\right)+6x\left(x+4\right)=0\)

\(\Leftrightarrow21x-6x^2+35-10x+6x^2+24x=0\)

\(\Leftrightarrow x=1\)

b: Ta có: \(x^3-25x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

a. (3x + 5)(7 - 2x) + 6x(x + 4) = 0

<=> 21x - 6x² + 35 - 10x + 6x² + 24x = 0

<=> -6x² + 6x² + 21x - 10x + 24x = -35

<=> 35x = -35

<=> x = \(\dfrac{-35}{35}=-1\)

b. x³ - 25x = 0

<=> x(x² - 5²)

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

<=> \(\left[{}\begin{matrix}x=0\\x+5=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\\x=5\end{matrix}\right.\)

a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)

\(\Leftrightarrow x\in\left\{-2;12\right\}\)

b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)

`@` `\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

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

<=> (x + 3)^2 - 4(x + 3) = 0

<=> (x + 3)(x - 1) = 0

<=> x = -3 hoặc x = 1

b) x(x + 5)(x- 5) - (x - 3)(x^2 + 3x + 9) = 7

<=> x^3 - 25x - x^3 + 27 = 7

<=> -25x + 27 = 7

<=> x = 4/5

a/ \(\left(x+3\right)^2-4x-12=0\)

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

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

\(\left[{}\begin{matrix}x+3=0\Rightarrow x=-3\\x+3-4=0\Rightarrow x=1\end{matrix}\right.\)

---

b/ \(x\left(x+5\right)\left(x-5\right)-\left(x-3\right)\left(x^2+3x+9\right)=7\)

\(x\left(x^2-25\right)-\left(x^3-27\right)=7\)

\(x^3-25x-x^3+27=7\)

\(-25x=-20\)

\(x=\dfrac{20}{25}=\dfrac{4}{5}\)

a, <=>x² +6x+9-4x-12=0

<=> x² +2x -3=0

<=> x² +3x -x-3=0

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

<=> (x-1)(x+3)=0

<=> x=1 hoặc x=-3

b, <=> x(x² -25) - (x-3)(x+3)² -7=0 

<=> x³ -25x + (9-x²) (x+3) -7=0

<=> x³ -25x+ 9x+27-x³ -3x² -7=0

<=> -3x² -16x +20=0

<=>(3x-10)(x-2) =0 (đoạn này tự phân tích nha ^ ^)

<=> x= 10/3 hoặc x=2

Chúc bạn học tốt nha!

a) `(x-8)(x^3+8)=0`

`<=>(x-8)(x+2)(x^2-2x+4)=0`

`<=>` \(\left[ \begin{array}{l}x=8\\x=-2\end{array} \right.\) (Vì `x^2-2x+4 \ne 0 forall x)`

Vậy `A={8;-2}`.

b) `(4x-3)-(x+5)=3(10-x)`

`,=>4x-3-x-5=30-3x`

`<=>3x-8=30-3x`

`<=>6x=38`

`<=>x=19/3`

Vậy `S={19/3}`.

a: Ta có: \(x^3+1=0\)

\(\Leftrightarrow x^3=-1\)

hay x=-1

b: Ta có: \(6x^2-12x-48=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)