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.

\(B=8x^3+12x^2+6x+1\)

\(=8\left(\dfrac{1}{2}\right)^3+12\left(\dfrac{1}{2}\right)^2+6.\dfrac{1}{2}+1\)

\(=8.\dfrac{1}{8}+12.\dfrac{1}{4}+3+1\)

\(=1+3+4\)

\(=8\)

Để tính giá trị của biểu thức B=8x^3+12x^2+6x+1 tại x=1/2, ta thay giá trị này vào biểu thức.

B = 8(1/2)^3 + 12(1/2)^2 + 6(1/2) + 1
= 8(1/8) + 12(1/4) + 6(1/2) + 1
= 1 + 3 + 3 + 1
= 8

Vậy, giá trị của biểu thức B tại x=1/2 là 8.

Thay \(x=\dfrac{1}{2}\) vào biểu thức trên , ta có : 

\(B=\)\(8.\left(\dfrac{1}{2}\right)^3+12.\left(\dfrac{1}{2}\right)^2+6.\dfrac{1}{2}+1\)

\(=8.\dfrac{1}{8}+12.\dfrac{1}{4}+6.\dfrac{1}{2}+1\)

\(=1+3+3+1\)

\(=4+4\)

\(=8\)

Vậy khi \(x=\dfrac{1}{2}\) thì \(B=8\)

\(2x-1^3+8\)

\(=2x-9\)

\(=\left(\sqrt{2x}\right)^2-3^2\)

\(=\left(\sqrt{2x}-3\right)\left(\sqrt{2x}+3\right)\)

_________

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

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

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

_______________

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

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

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

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

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

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

________

\(9x^3-12x^2+6x-1\)

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

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

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

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

b: 8x^3-12x^2+6x-1

=(2x)^3-3*(2x)^2*1+3*2x*1^2-1^3

=(2x-1)^3

c: =(8x^3-12x^2+6x-1)-1

=(2x-1)^3-1

=(2x-1-1)[(2x-1)^2+2x-1+1]

=2(x-1)(4x^2-4x+1+2x)

=2(x-1)(4x^2-2x+1)

8x³ - 12x² + 6x - 1

= (2x)³ - 3.(2x)².1 + 3.2x.1 - 1³

= (2x - 1)³

--------------------

8x³ - 12x² + 6x - 2

= 8x³ - 12x² + 6x - 1 - 1

= (2x)³ - 3.(2x)².1 + 3.(2x).1 - 1³ - 1³

= (2x - 1)³ - 1³

= (2x - 1 - 1)[(2x - 1)² + (2x - 1).1 + 1]

= (2x - 2)(4x² - 4x + 1 + 2x - 1 + 1)

= 2(x - 1)(4x² - 2x + 1)

--------------------

9x³ - 12x² + 6x - 1

= x³ + 8x³ - 12x² + 6x - 1

= x³ + (2x)³ - 3.(2x)² + 3.2x.1² - 1³

= x³ + (2x - 1)³

= (x + 2x - 1)[x² - x.(2x - 1) + (2x - 1)²]

= (3x - 1)(x² - 2x² + x + 4x² - 4x + 1)

= (3x - 1)(3x² - 3x + 1)

\(72x^2-60x+18x-15-72x^2+16x-27x+6=203\)

\(-53x-9=203\)

\(-53x=212\)

\(x=\frac{106}{27}\)

tìm nghiệm nha mọi người quên chưa nói

a) \(x^2+2x+3=0\)

\(\Rightarrow x^2+2x+3-3=0-3\)

\(\Rightarrow x^2+2x=-3\)

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

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

Điều này là vô lý vì bình phương của 1 số luôn lớn hơn hoặc bằng 0 mà -2 < 0.

Vậy đa thức vô nghiệm.

b) \(x^2+6x+5=0\)

\(x^2+x+5x+5=0\)

\(x\left(x+1\right)+5\left(x+1\right)=0\)

\(\left(x+5\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+5=0\\x+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-5\\x=-1\end{cases}}\)

Ta có: \(x^3+12x^2+48x+64=8x^3-12x^2+6x-1\)

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

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

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

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

\(\Leftrightarrow\left(3-x\right)\left(7x^2+3x-6\right)=0\)

\(\Leftrightarrow7\left(3-x\right)\cdot\left(x^2+\frac{3}{7}x-\frac{6}{7}\right)=0\)

mà 7>0

nên \(\left[{}\begin{matrix}3-x=0\\x^2+\frac{3}{7}x-\frac{6}{7}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2+2\cdot x\cdot\frac{3}{14}+\frac{9}{196}-\frac{177}{196}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\left(x+\frac{3}{14}\right)^2=\frac{177}{196}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+\frac{3}{14}=\frac{\sqrt{177}}{14}\\x+\frac{3}{14}=-\frac{\sqrt{177}}{14}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-3+\sqrt{177}}{14}\\x=\frac{-3-\sqrt{177}}{14}\end{matrix}\right.\)

Vậy: \(S=\left\{3;\frac{-3+\sqrt{177}}{14};\frac{-3-\sqrt{177}}{14}\right\}\)

x³ + 12x² + 48x + 64 = 8x³ - 12x² + 6x - 1

\(\Leftrightarrow\) x³ + 12x² + 48x + 64 - 8x³ + 12x² - 6x + 1 = 0

\(\Leftrightarrow\) -7x³ + 24x² + 42x + 65 = 0

Bn cho đề thế này ai mà giải được :vvv