d) <=>x²-5x-x+5=0

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

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

<=>x=5 hoặc x=1

thank nha

a) (x-2)³ - 6(x+1)² - x³ + 12 = 0 

<=> x³-6x²+12x-8-6(x²+2x+1)-x³+12=0

<=> x³-6x²+12x-8-6x²-12x-6-x³+12=0

<=> -12x²+4=0

<=> \(x=\frac{1}{\sqrt{3}},x=-\frac{1}{\sqrt{3}}\)

vậy pt có 2 nghiệm....

b) x³ - 6x² + 12x - 8 = 0 

<=> (x³-2x²)-(4x²-8x)+(4x+8)=0

<=> (x-2)(x²-4x+4)=(x-2)³=0

=> x=2 là nghiệm

c) 8x³ - 12x² + 6x - 1 = 0

<=> (2x-1)³=0

<=> x=1/2

a) \(\left(x-2\right)^3-6\left(x+1\right)^2-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-6\left(x^2+2x+1\right)-x^3+12=0\)

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

\(\Leftrightarrow-12x^2-2=0\)

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

\(\Leftrightarrow6x^2+1=0\) (vô nghiệm)

Vậy không có giá trị nào của x thỏa mãn pt

b) \(x^3-6x^2+12x-8=0\)

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

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy x=2

c) \(8x^3-12x^2+6x-1=0\)

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

\(\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)

Vậy \(=\frac{1}{2}\)

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

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

\(-4x^2+12x-8-6x^2-12x+6=0\)

-10x^2-2=0

5x^2+1=0

x^2=-1/5

x=\(\varnothing\)

\(x^3+6x^2+12x+8=0\)

\(\Rightarrow x^3+8+6x^2+12x=0\)

\(\Rightarrow x^3+2^3+3.2.x^2+3.2^2.x=0\)

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

\(\Rightarrow x+2=0\Rightarrow x=-2\)

c: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

=(x - 2)³ = 0

x = 2

x=0 hoặc x=2/3

x=2 đó bạn

a: \(x^2+12x+36=0\) 

=>\(x^2+2\cdot x\cdot6+6^2=0\)

=>\(\left(x+6\right)^2=0\)

=>x+6=0

=>x=-6

b: \(4x^2-4x+1=0\)

=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)

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

=>2x-1=0

=>2x=1

=>x=1/2

c: \(x^3+6x^2+12x+8=0\)

=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)

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

=>x+2=0

=>x=-2

(x+5)³-x³-125

=x³+5³-x³-5³

=0

1. \(\left(x+5\right)^3-x^3-125\)

\(=x^3+15x^2+75x+125-x^3-125\)

\(=15x^2+75x\)

2. \(x^3+6x^2+12x+8=0\)

\(\Leftrightarrow x^3+2x^2+4x^2+8x+4x+8=0\)

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

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

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

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

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

1. \(\left(x+5\right)^3-x^3-125\)  

=\(x^3+3\cdot x^2\cdot5+3\cdot x\cdot5^2+5^3-x^3-125\)  

=\(x^3+15x^2+75x+125-x^3-125\)  

=\(15x^2+75x\)   

=\(15x\left(x+5\right)\)      

2. \(x^3+6x^2+12x+8=0\)   

\(x^3+2x^2+4x^2+8x+4x+8=0\)   

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

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

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

 \(\left(x+2\right)^3=0\)    

\(x+2=0\) 

\(x=-2\)

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