1) \(8x^3-12x^2+6x-1=0\)

\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)

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

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

2) \(x^3-6x^2+12x-8=27\)

\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)

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

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

\(\Leftrightarrow x-2=3\)

\(\Leftrightarrow x=3+2\)

\(\Leftrightarrow x=5\)

3) \(x^2-8x+16=5\left(4-x\right)^3\)

\(\Leftrightarrow\left(x-4\right)^2=5\left(4-x\right)^3\)

\(\Leftrightarrow\left(4-x\right)^2=5\left(4-x\right)^3\)

\(\Leftrightarrow5\left(4-x\right)=1\)

\(\Leftrightarrow4-x=\dfrac{1}{5}\)

\(\Leftrightarrow x=4-\dfrac{1}{5}\)

\(\Leftrightarrow x=\dfrac{19}{5}\)

4) \(\left(2-x\right)^3=6x\left(x-2\right)\)

\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)

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

\(\Leftrightarrow8-x^3=0\)

\(\Leftrightarrow x^3=8\)

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

\(\Leftrightarrow x=2\)

5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)

\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)

\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)

\(\Leftrightarrow12x-4=-10\)

\(\Leftrightarrow12x=-10+4\)

\(\Leftrightarrow12x=-6\)

\(\Leftrightarrow x=\dfrac{-6}{12}\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)

\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)

\(\Leftrightarrow-54x-2x^3=36x^2-54x\)

\(\Leftrightarrow-2x^3=36x^2\)

\(\Leftrightarrow-2x^3-36x^2=0\)

\(\Leftrightarrow-2x^2\left(x+18\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)

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.

Bài 4:

a, \(x^3+12x^2+48x+64=x^3+4x^2+8x^2+32x+16x+64\)

\(=x^2.\left(x+4\right)+8x.\left(x+4\right)+16.\left(x+4\right)\)

\(=\left(x+4\right).\left(x^2+8x+16\right)=\left(x+4\right).\left(x^2+4x+4x+16\right)\)

\(=\left(x+4\right).\left(x+4\right)^2=\left(x+4\right)^3\)(1)

Thay \(x=6\) vào (1) ta được:

\(\left(6+4\right)^3=10^3=1000\)

Vậy...........

b, \(x^3-6x^2+12x-8=x^3-2x^2-4x^2+8x+4x-8\)

\(=x^2.\left(x-2\right)-4x.\left(x-2\right)+4.\left(x-2\right)\)

\(=\left(x-2\right).\left(x^2-4x+4\right)=\left(x-2\right).\left(x^2-2x-2x+4\right)\)

\(=\left(x-2\right).\left(x-2\right)^2=\left(x-2\right)^3\)(2)

Thay \(x=22\) vào (2) ta được:

\(\left(22-2\right)^3=20^3=8000\)

Vậy.............

Chúc bạn học tốt!!!

Bài 2:

a, \(\left(x+9\right)^3=27=3^3\)

\(\Rightarrow x+9=3\Rightarrow x=-6\)

Vậy.........

b, \(8-12x-x^3+6x^2=-64\)

\(\Rightarrow-\left(x^3-6x^2+12x-8\right)=-64\)

\(\Rightarrow x^3-2x^2-4x^2+8x+4x-8=64\)

\(\Rightarrow x^2.\left(x-2\right)-4x.\left(x-2\right)+4.\left(x-2\right)=64\)

\(\Rightarrow\left(x-2\right).\left(x^2-4x+4\right)=64\)

\(\Rightarrow\left(x-2\right).\left(x^2-2x-2x+4\right)=64\)

\(\Rightarrow\left(x-2\right).\left(x-2\right)^2=64\)

\(\Rightarrow\left(x-2\right)^3=4^3\Rightarrow x-2=4\Rightarrow x=6\)

Vậy............

Chúc bạn học tốt!!!

giup minh voi dang can

a, <=> x² -2x +1 + 5x -x² =8

<=> 3x +1 =8 

<=> 3x = 7

<=> x= 7/3

b, thiếu đề

c, <=> 2x³ -1 + 2x(4 -x²) = 7

<=> 2x³ + 8x -2³ = 8

<=> 8x =8

<=> x=1

\(\Leftrightarrow x\cdot\dfrac{5}{2}-3x^2+15-18x+12x-2+\dfrac{1}{4}x+3=0\)

\(\Leftrightarrow-3x^2-\dfrac{13}{4}x+16=0\)

\(\Leftrightarrow-12x^2-13x+64=0\)

hay \(x\in\left\{1,83;-2,91\right\}\)