mày viết lại cái đề bài hộ tao cái

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

=>(x+3)(x^2+6x+9)=0

=>x=-3

b: \(\Leftrightarrow x^2-2x-x^3+6x^2-4=0\)

=>-x^3+6x^2-2x-4=0

hay \(x\in\left\{5.5;1.14;-0.64\right\}\)

c: =>(3x-1)^3=8

=>3x-1=2

=>3x=3

=>x=1

(x+1)(x-2)(x-3)(2x-1)(3x-1)=0

\(\Leftrightarrow\) \(6x^5-12x^4-17x^4+34x^3-7x^3+14x^2+13x^2-26x-3x+\)6 =0

\(6x^5-29x^4+27x^3+27x^2-29x+6=0\)

\(\Leftrightarrow\left(6x^5-18x^4\right)+\left(-11x^4+33x^3\right)+\left(-6x^3+18x^2\right)+\left(9x^2-27x\right)+\left(-2x+6\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(6x^4-11x^3-6x^2+9x-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\left(6x^4-12x^3\right)+\left(x^3-2x^2\right)+\left(-4x^2+8x\right)+\left(x-2\right)\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(6x^3+x^2-4x+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(\left(6x^3+6x^2\right)+\left(-5x^2-5x\right)+\left(x+1\right)\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(6x^2-5x+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(\left(6x^2-3x\right)+\left(-2x+1\right)\right)=0\)

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

\(\Leftrightarrow x=\left(3;2;-1;\frac{1}{2};\frac{1}{3}\right)\)

\(1-27x^3\)

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

\(=\left(1-3x\right)\left(1+3x+9x^2\right)\)

\(---\)

\(x-3^3+27\)

\(=x-27+27=x\) 

\(---\)

\(27x^3+27x^2+9x+1\)

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

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

\(---\)

\(\dfrac{x^6}{27}-\dfrac{x^4y}{3}+x^2y^2-y^3\) (sửa đề)

\(=\left(\dfrac{x^2}{3}\right)^3-3\cdot\left(\dfrac{x^2}{3}\right)^2\cdot y+3\cdot\dfrac{x^2}{3}\cdot y^2-y^3\)

 \(=\left(\dfrac{x^2}{3}-y\right)^3\)

#Ayumu

1-27x\(^3\)

=(1-3x)(1+3x+9x\(^2\)

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/ \(x^3-7x+6=0\)

\(\Leftrightarrow x^3+3x^2-3x^2-9x+2x+6=0\)

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

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

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

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

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

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

hoặc   \(x-1=0\)

hoặc   \(x+2=0\)

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

hoặc   \(x=1\)

hoặc   \(x=-2\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-3;1;-2\right\}\)

2/ \(x^3-6x^2-x+30\)

\(\Leftrightarrow x^3+2x^2-8x^2-16x+15x+30=0\)

\(\Leftrightarrow x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-8x+15\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-3x-5x+15\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[x\left(x-3\right)-5\left(x-3\right)\right]=0\)

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

\(\Leftrightarrow\)\(x+2=0\)

hoặc   \(x-3=0\)

hoặc   \(x-5=0\)

\(\Leftrightarrow\)\(x=-2\)

hoặc   \(x=3\)

hoặc   \(x=5\)

Vậy tập nghiệm của phương trình là :\(S=\left\{-2;3;5\right\}\)

3/ \(x^3-9x^2+6x+16=0\)

\(\Leftrightarrow x^3+x^2-10x^2-10x+16x+16=0\)

\(\Leftrightarrow x^2\left(x+1\right)-10x\left(x+1\right)+16\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-10x+16\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-8x-2x+16\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[x\left(x-8\right)-2\left(x-8\right)\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-8\right)\left(x-2\right)=0\)

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

hoặc  \(x-8=0\)

hoặc  \(x-2=0\)

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

hoặc   \(x=8\)

hoặc   \(x=2\)

Vậy tập nghiệm của phương trình là :\(S=\left\{-1;8;2\right\}\)

4/ Đề bài sai ! Sửa lại nhé :

 \(2x^3-x^2+5x+3=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x^2-x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(tm\right)\\\left(x-\frac{1}{2}\right)^2+\frac{11}{4}=0\left(ktm\right)\end{cases}}\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-\frac{1}{2}\right\}\)

a, \(4x^2-4x=-1\Leftrightarrow4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\frac{1}{2}\)

b, \(27x^3+27x^2+9x+1=0\Leftrightarrow27x^3+1+27x^2+9x=0\)

\(\Leftrightarrow\left(3x+1\right)\left(9x^2-3x+1\right)+9x\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(9x^2+2>0\right)=0\Leftrightarrow x=-\frac{1}{3}\)

c, \(9x^2\left(x+1\right)-4\left(x+1\right)=0\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\Leftrightarrow x=-\frac{2}{3};x=\frac{2}{3};x=-1\)

d, \(\left(x+1\right)^3-25\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)^2-25\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-4\right)\left(x+6\right)=0\Leftrightarrow x=-1;x=-6;x=4\)