a) \(x^3-6x^2+12x-9=0\)

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

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

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

b) \(8x^3+12x^2+6x-26=0\)

\(\Leftrightarrow8x^3+12x^2+6x+1-27=0\)

\(\Leftrightarrow\left(2x+1\right)^3=27\)

\(\Leftrightarrow2x+1=3\Leftrightarrow x=1\)

x^3-6^2+12x-8=1

(x-2)^3=1

=>x-2=1

=>x=3

Câu b tương tự nha

Ta có : \(x^2-2x-1=0 \)
\(\Leftrightarrow \)\((x-1)^2=2\)
\(\Leftrightarrow \)\(\left[\begin{array}{} x-1=\sqrt{2}\\ x-1=-\sqrt{2} \end{array} \right.\)
Đặt P = \(\dfrac{x^6-6x^5+12x^4-8x^3+2015}{x^6-8x^3-12x^2+6x+2015}\)
          =\(\dfrac{(x^6-2x^5-x^4)-(4x^5-8x^4-4x^3)+(5x^4-10x^3-5x^2)-(2x^3-4x^2-2x)+(x^2-2x-1)+2016} {(x^6-2x^5-x^4)+(2x^5-4x^4-2x^3)+(5x^4-10x^3-5x^2)+(4x^3-8x^2-4x)+(x^2-2x-1)+12x+2016}\)
         =\(\dfrac{x^4(x^2-2x-1)-4x^3(x^2-2x-1)+5x^2(x^2-2x-1)-2x(x^2-2x-1)+(x^2-2x-1)+2016} {x^4(x^2-2x-1)+2x^3(x^2-2x-1)+5x^2(x^2-2x-1)+4x(x^2-2x-1)+(x^2-2x-1)+12x+2016}\)
         =\(\dfrac{2016}{12x + 2016}\)
         =\(\dfrac{2016}{12(x+1)+2004}\)
         =\(\dfrac{168}{x+1+167}\)
         =\(\left[\begin{array}{} \dfrac{168}{\sqrt{2}+167}\\ \dfrac{168}{-\sqrt{2}+167} \end{array} \right.\)
Chú thích: Hình như mẫu là \(-6x\) chứ không phải \(6x \) bạn ạ. Hay là mình phân tích sai thì cho mình xin lỗi nhé.

8x^3 - 8x^2 + 20x^2 - 20x + 26x - 26 =0

<=> 8x^2 ( x-1) + 20x(x-1) + 26(x-1)=0

<=>( 8x^2 + 20x + 26)(x-1)=0

<=> ( x-1)= 0 ( Vì 8x^2 + 20x + 26 >=13,5)

<=> x=1

\(\Leftrightarrow\left(2x\right)^3-3.\left(2x\right)^2.1+3.2x.1^2-1=8\Leftrightarrow\left(2x-1\right)^3=2^3\)

\(\Leftrightarrow2x-1=2\Rightarrow2x=\frac{3}{2}\Rightarrow\frac{3}{4}\)

=> x = 3/4 viết thiếu nhe

a: \(x^4=5x^2+2x-3\)

=>\(x^4-5x^2-2x+3=0\)

=>\(x^4+x^3-x^2-x^3-x^2+x-3x^2-3x+3=0\)

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

TH1: \(x^2+x-1=0\)

=>\(x^2+x+\frac14=\frac54\)

=>\(\left(x+\frac12\right)^2=\frac54\)

=>\(x+\frac12=\pm\frac{\sqrt5}{2}\)

=>\(x=-\frac12\pm\frac{\sqrt5}{2}\)

TH2: \(x^2-x-3=0\)

=>\(x^2-x+\frac14-\frac{13}{4}=0\)

=>\(\left(x-\frac12\right)^2=\frac{13}{4}\)

=>\(x-\frac12=\pm\frac{\sqrt{13}}{2}\)

=>\(x=\frac12\pm\frac{\sqrt{13}}{2}\)

c: \(3x^3+3x^2+3x=-1\)

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

=>\(\left(x+1\right)^3=\left(x\cdot\sqrt[3]{-2}\right)^3\)

=>\(x+1=x\cdot\sqrt[3]{-2}\)

=>\(x\left(1-\sqrt[3]{-2}\right)=-1\)

=>\(x=\frac{-1}{1-\sqrt[3]{-2}}\)

d: \(8x^3-12x^2+6x-5=0\)

=>\(8x^3-12x^2+6x-1-4=0\)

=>\(\left(2x-1\right)^3=4\)

=>\(2x-1=\sqrt[3]{4}\)

=>\(2x=1+\sqrt[3]{4}\)

=>\(x=\frac12+\frac12\cdot\sqrt[3]{4}\)