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

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

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

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

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

a) x^3 - 6x^2 + 12x -8 = 0
x^3 - 3.x^2 .2 + 3.x.2^2 - 2^3 = 0
=> ( x-2) = 0
=> x-2=0 <=> x=2

b) 8x^3 - 12x^2 + 6x -1 = 0
(2x)^3 - 3.(2x)^2.1 + 3.2x.1 -1^3 = 0
=> ( 2x - 1 ) = 0
=> 2x-1 = 0 <=> 2x = 1
x = 1/2

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.

A= x^3-3x^2+3x5 

=x²(3x³+x-3)

Để giá trị của A nhỏ nhất 

=>x=2.Thay x=2 vào ta đc:

A=2²(3*2³+2-3)=4(3*8+2-3)

=4(24+2-3)=4*23=92

B=x^3 + 6x^2+12x-1   

=x³+6x²+12x+8-9

=(x+2)³-9

Để giá trị của B nhỏ nhất 

=>x=-1.Thay x=-1 vào ta được:

B=[(-1)+2]³-9=[1]³-9=-8

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

<=>  \(4x^3+6x^2+3x-13=0\)

<=>  \(4x^3-4x^2+10x^2-10x+13x-13=0\)

<=>  \(4x^2\left(x-1\right)+10x\left(x-1\right)+13\left(x-1\right)=0\)

<=>  \(\left(x-1\right)\left(4x^2+10x+13\right)=0\)

<=> \(x-1=0\)

<=>  \(x=1\)

Vậy...

Cam on ban nhieu nhe <3

\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a)      \(2\left(x+5\right)-x^2-5x=0\)

  \(\Leftrightarrow2x+10-x^2-5x=0\)

 \(\Leftrightarrow-x^2-3x+10=0\)

\(\Leftrightarrow x^2+3x-10=0\)

 \(\Leftrightarrow x^2-2x+5x-10=0\)

\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)

b) \(x^3-6x^2+12x-8=0\)

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

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

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

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

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

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

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

c)\(16x^2-9\left(x+1\right)^2=0\)

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

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

\(\Leftrightarrow\left(x-1\right)\left(7x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\7x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)

d) \(x^3+x=0\)

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

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

e)\(x^2-2x-3=0\)

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

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

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

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

Cảm ơn bạn nha