Bài 1:

a) Ta có: \(x^3+3x^2+3x+2=0\)

\(\Leftrightarrow x^3+2x^2+x^2+2x+x+2=0\)

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

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

mà \(x^2+x+1>0\forall x\)

nên x+2=0

hay x=-2

Vậy: x=-2

b) Ta có: \(x^3-12x^2+48x-72=0\)

\(\Leftrightarrow x^3-6x^2-6x^2+36x+12x-72=0\)

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

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

mà \(x^2-6x+12>0\forall x\)

nên x-6=0

hay x=6

Vậy: x=6

a) Ta có: \(x^3+3x^2+3x+2=0\)

\(\Leftrightarrow x^3+2x^2+x^2+2x+x+2=0\)

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

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

mà \(x^2+x+1\ne0\forall x\)

nên x+2=0

hay x=-2

Vậy: x=-2

b) Ta có: \(x^3-12x^2+48x-72=0\)

\(\Leftrightarrow x^3-6x^2-6x^2+36x+12x-72=0\)

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

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

mà \(x^2-6x+12\ne0\forall x\)

nên x-6=0

hay x=6

Vậy: x=6

a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0

=>-5x-4=0

=>x=-4/5

b: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81

=>83x=83

=>x=1

Cảm ơn nhìu ạ :3

câu c,d,e lạ quá nhưng thui you viết thế nào tui làm thế ấy sai đừng trách

 a) 5x(x-1)=x-1

<=> 5x(x-1)-(x-1)=0

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

<=>x-1=0 hoặc 5x-1=0

<=>x=1 hoặc \(\frac{1}{5}\)

b) 2(x+5)-x*2-5x=0

VT=-5(x-2)

<=>-5(x-2)=0

<=>x-2=0

<=>x=2

c)(2x-3)*2-(x-5)*2=0

VT=2(x+2)

<=>2(x+2)=0

<=>x+2=0

<=>x=-2

d) 3x*3-48x=0

VT=-39x

<=>-39x=0

<=>x=0

e) x*3+x*2-4x=4

VT=x

<=>x=4

\(\left|2x-\frac{1}{2}\right|+1=3x\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}-x=-\frac{1}{2}\\5x=\frac{3}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{10}\end{cases}}\)

1) <=> x² - 4x - x² + 8 = 0 <=> x² - 4x + 8 =  0 

Dễ thấy phương trình vô nghiệm vì x² - 4x + 8 = ( x - 2 )² + 4 > 0

2) <=> ( x - 1 )³ = 0 <=> x = 1

3) <=> ( x - 2 )³ = 0 <=> x = 2

4) <=> ( 2x - 1 )³ = 0 <=> x = 1/2

\(1,\)

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

\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)

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

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

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

\(2,\)

\(3x\left(x+5\right)-6\left(x+5\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

\(3,\)

\(x^4-x^2=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

\(4,\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(5,\)

\(x\left(x+6\right)-10\left(x-6\right)=0\)

\(\Leftrightarrow x^2+6x-10x+60=0\)

\(\Leftrightarrow x^2-4x+60=0\)

\(\Leftrightarrow x^2-4x+4+56=0\)

\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)

=> Phương trình vô nghiệm

a) \(x^3-9x^2+27x-27=-8\Leftrightarrow\left(x-3\right)^3=\left(-2\right)^3\)

\(\Rightarrow x-3=-2\Leftrightarrow x=1\) vậy \(x=1\)

b) \(64x^3+48x^2+12x+1=0\Leftrightarrow\left(4x+1\right)^3=3^3\)

\(\Rightarrow4x+1=3\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{2}{4}=\dfrac{1}{2}\) vậy \(x=\dfrac{1}{2}\)

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

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\) vậy \(x=1\)

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

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

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

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\) vậy \(x=2\)