a) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=27\)

\(\Rightarrow x^3+3^3-x\left(x^2-4\right)=27\)

\(\Rightarrow x^3+27-x^3+4x=27\)

\(\Rightarrow27+4x=27\)

\(\Rightarrow4x=0\)

\(\Rightarrow x=0\)

b) \(2x^2+7x+3=0\)

\(\Rightarrow2x^2+x+6x+3=0\)

\(\Rightarrow x\left(2x+1\right)+3\left(2x+1\right)=0\)

\(\Rightarrow\left(2x+1\right)\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=-1\\x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

c) Trùng đề bài a

d) \(2x^2+11x+9=0\)

\(\Rightarrow2x^2+2x+9x+9=0\)

\(\Rightarrow2x\left(x+1\right)+9\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(2x+9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\2x+9=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\2x=-9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{9}{2}\end{matrix}\right.\)

a) \(4x^2-12x=-9\)

\(\Leftrightarrow4x^2-12x+9=0\)

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

\(\Leftrightarrow2x-3=0\Leftrightarrow x=\frac{3}{2}\)

b) \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(25-4x^2\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(5+2x\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7+5+2x\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(4x+12\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-3\end{array}\right.\)

c)\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=0\\x=2\end{array}\right.\)

d) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right]\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\x=-\frac{23}{17}\end{array}\right.\)

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)

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

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

`<=>[(x=2/3),(x=-5):}`

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

`<=>(5x-1)(x-3)(x+3)=0`

`<=>[(x=1/5),(x=3),(x=-3):}`

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

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

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

`13)x(x-5)-4x+20=0`

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

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

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

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

`<=>[(x=-5),(x=1):}`

\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

b) Ta có: \(\left(5-x\right)^3+27=0\)

\(\Leftrightarrow\left(5-x\right)^3=-27\)

\(\Leftrightarrow5-x=-3\)

hay x=8

d) Ta có: \(\left(x^2-1\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=-7\end{matrix}\right.\)

f) Ta có: \(\left(x^2+81\right)\left(x-7\right)\left(x^2-2\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

x^3+27+(x+3).(x-9)=0

<=> x^3+3^3+(x+3).(x+9)=0

<=> (x+3).(x^2-3x+9)+(x+3).(x-9)=0

<=> (x+3).[(x^2 -3x+9)+(x-9)]=0

<=> (x+3).(x^2- 3x+9+x-9)=0

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

<=> x+3=0

hay x^2-2x=0

<=> x=-3

hay x=2

Vậy, x=-3; x=2

\(2x^2-x-6=0\)

\(\Leftrightarrow2x^2+3x-4x-6=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=2\end{matrix}\right.\)

Vậy .................

x = từ 1 đến 10000....0

a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)

f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)

g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)

i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

⇔ \(x^3+27+x^2-9x+3x-27=0\)

⇔\(x^3+x^2-6x=0\)

⇔\(x\left(x^2+x-6\right)=0\)

⇔\(x=0\) hoặc \(x^2+x-6=0\)

⇔\(x=0\) hoặc \(x^2+x=6\)

⇔ x=0 hoặc x(x+1)=6

⇔ x=0 hoặc x=6 và x=5

Vậy x={0;5;6}

⇔ (x + 3)³- 9x( x + 3) + (x + 3)( x - 9)=0

⇔ (x + 3)³ - [ 9x( x + 3) - (x + 3)( x - 9)] =0

⇔ (x + 3)³- [( x + 3)(8x + 9)] = 0

⇔ x³ + 27x + 9x² + 27 - 8x² - 9x - 24x - 27

⇔ x³ + x² - 6x = 0

⇔ x( x² + x - 6) = 0

⇔ x( x - 2 )( x + 3) = 0

⇔ \(\left[{}\begin{matrix}x=0\\x-2=0\\x+3=0\end{matrix}\right.\text{⇔}\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)

a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)

b: =>x-1/2=1/3

=>x=5/6

c: =>2/3x-1=0 hoặc 3/4x+1/2=0

=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3

d =>4/9:x=10/3:9/4=10/3*4/9=40/27

=>x=4/9:40/27=4/9*27/40=108/360=3/10