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

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

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

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

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

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

<=> (x+3)(x²+3x+9)+(x+3)(x - 9)

<=> (x+3)(x²-3x+9+x - 9)=0

<=> (x+3)(x²-2x)=0

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

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

\(x^3+27+\left(x+3\right)\left(x-9\right)\)=0
\(\Leftrightarrow x^3+3^3+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x+3=0\\x-2=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-3\\x=2\end{array}\right.\)
Vậy x=0 hoặc x=-3 hoặc x=2

Đáp án của bạn Nguyễn Tiến Hải còn thiếu trường hợp:

(x + 1/2)² = 25/4

TH1: x + 1/2 = 5/2 và giải như bạn Hải

TH2: x + 1/2 = -5/2

         x = -3

=> (x+3)(x²-3x+9) + (x+3)(x-9) =0
=> (x+3)(x²-2x)=0 => (x+3)(x-2)x=0
=> x=-3 hoặc x=2 hoặc x=0

a) \(x+546=46\\ x=46-546\\ x=-500\)

b) \(2x-19\times3=27\\ 2x-57=27\\ 2x=27+57\\ 2x=84\\ x=84:2\\ x=42\)

c) \(x+12=23+3\times3^4\\ x+12=23+3\times81\\ x=23+243-12\\ x=254\)

d) \(x-12=3-3\times2^4\\ x-12=3-3\times16\\ x=3-48+12\\ x=-33\)

e) \(\left(27-x\right)\left(x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}27-x=0\\x+9=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=27\\x=-9\end{matrix}\right.\)

f) \(\left(-x\right)\left(x-43\right)=0\\ \Rightarrow\left[{}\begin{matrix}-x=0\\x-43=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=43\end{matrix}\right.\)

a) x + 546 = 46

x = 46 - 546

x = -500

b) 2x - 19 x 3 = 27

2x - 57 = 27

2x = 27 + 57

2x = 84

x = 84 : 2

x = 42

c) x + 12 = 23 + 3 x 3⁴

x + 12 = 23 + 3 x 81

x + 12 = 23 + 243

x + 12 = 266

x = 266 - 12

x = 254

d) x - 12 = 3 - 3 x 2⁴

x - 12 = 3 - 3 x 16

x - 12 = 3 - 48

x - 12 = -45

x = -45 + 12

x = -33

e) (27 - x) x (x + 9) = 0

TH1: 27 - x = 0

x = 27 - 0

x = 27

TH2: x + 9 = 0

x = 0 + 9

x = 9

⇒ x = 27 hoặc x = 9

f) (-x) x (x - 43) = 0

TH1: -x = 0

⇒ x = 0

TH2: x - 43 = 0

x = 0 + 43

x = 43

⇒ x = 0 hoặc x = 43.

x³+27+(x+3).(x-9)=0 nha mk ghi nhầm 

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

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

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

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

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

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

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

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

b: Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-\frac{2}{3}\end{cases}}}\)

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

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

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

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

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

\(\Leftrightarrow\hept{\begin{cases}x=0\\x+3=0\\x-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-3\\x=2\end{cases}}}\)

b) \(x^2-2x-3=0\)

\(D=b^2-4ac\)

\(\left(-2\right)^2-\left(4\left(1.3\right)\right)=16\)

\(x_{1,2}=\frac{-b-\sqrt{D}}{2a}=\frac{2-\sqrt{16}}{2}\)

\(x=1;-3\)

a)1/3
b)-1
c)3

\(\frac{x^3\left(-1\right)}{9}=-\frac{x^3}{9}\)

\(-\frac{x^3}{9}=0\Rightarrow x=0\)

1) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

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

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

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

\(\Leftrightarrow-4x+10=0\)

\(\Leftrightarrow-4x=-10\)

\(\Leftrightarrow x=\frac{5}{2}.\)

Vậy \(S=\left\{\frac{5}{2}\right\}\)

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

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

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

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

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

\(\Leftrightarrow x+3=0\)hoặc \(x=0\)hoặc \(x-2=0\)

\(\Leftrightarrow x=-3\)hoặc \(x=0\)hoặc \(x=2\)

Vậy \(S=\left\{-3;0;2\right\}\)