3x³ - 27 x = 0

=> 3x ( x² - 9 ) = 0

\(=>\orbr{\begin{cases}3x=0\\x^2-9=0\end{cases}}\) 

\(=>\orbr{\begin{cases}x=0\\x^2=9\end{cases}}\)

=> x = + 3

hoặc x = 0

\(3x^2-27x=0\)

\(\Leftrightarrow3x\left(x+3\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}\)

c: =>(x-1)(x+1)=0

hay \(x\in\left\{1;-1\right\}\)

plss

a,

\(=\dfrac{3}{4x}.\left(x-3\right)\left(x+3\right)\)=0

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

=>\(x=\left\{3,-3\right\}\)

b,

\(x^3-16x=0\\x\left(x^2-16\right)\\ x\left(x-4\right)\left(x+4\right)\)

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

=>\(x=\left\{-4,0,4\right\}\)

d,

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

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

=>\(x=\left\{-3,0,3\right\}\)

e,

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

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

=>\(x=\left\{-2,-1,0\right\}\)

f,

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

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

a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x+25=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

b) Ta có: \(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

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

c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

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

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

d) Ta có: \(x^3-x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

e) Ta có: \(27x^3-27x^2+9x-1=1\)

\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)

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

\(\Leftrightarrow3x-1=1\)

\(\Leftrightarrow3x=2\)

hay \(x=\dfrac{2}{3}\)

1)3.x^2 - 75 = 0

3.x^2 - 3.25 = 0

3.(x^2-25)=0

x^2-5^2=0

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

=> x-5=0 hoặc x+5=0

=> x=5 hoặc x=-5

1) \(3x^2-75=0\)

\(\Leftrightarrow3\left(x^2-25\right)=0\)

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

\(\Leftrightarrow x^2=25\)

\(\Leftrightarrow x=\pm\sqrt{25}=\pm5\)

2) \(x^3+9x^2+27x+27=0\)

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

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

3) \(x^3+3x^2+3x=0\)

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

\(\Leftrightarrow\left(x+1\right)^3=1^3\)

\(\Leftrightarrow x+1=1\Leftrightarrow x=0\)

2, x^3 + 9x^2 +27x + 27 = 0

x^3 + 3.x^2.3 + 3.x.3^2 +3^3 = 0

(x+3)^3=0

x+3=0

x=-3

a, 3x^2-27x=0

3x(x-9)=0

3x=0=>x=0

x-9=0=>x=9

b,2/3x(x^2-4)=0

2/3x=0=>x=0

x^2-4=0=>x=2

\(3x^2-27x=0\)

\(3x\left(x-9\right)=0\)

\(\left[\begin{array}{nghiempt}x=0\\x-9=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=0\\x=9\end{array}\right.\)

\(\frac{2}{3}x\left(x^2-4\right)=0\)

\(\frac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)

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

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

a)\(3x^2-27x=0\)

\(3x\left(x-9\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}3x=0\\x-9=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=9\end{array}\right.\)

b) \(\frac{2}{3}x\left(x^2-4\right)=0\)

\(\frac{2}{3}x\left(x+2\right)\left(x-2\right)=0\)

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

3x³ - 27x = 0

=> 3x.(x² -9) = 0

=> 3x = 0 hoặc x² - 9 = 0

TH1: 3x = 0

=> x = 0

TH2: x² - 9 = 0

=> x² = 9

=> x = 3 hoặc x = -3

Vậy, x \(\in\){ 0; 3; -3}

\(3x^3-27x=0\)

\(3x.x^2-27x=0\)

\(3x.x^2=27x\)

\(x^2=27x:3x\)

\(x^2=9\)

\(x=3\).Vậy x = 3

Em học lớp 7 lận

em nói thật đấy

3x³ - 27x = 0

<=> x( 3x² - 27 ) = 0

=> x = 0 hoặc 3x² - 27 = 0

=> x = 0 hoặc 3x² = 27

=> x = 0 hoặc x² = 9 

=> x = 0 hoặc x = + - 3