a,4x^2-4x+1=0

  4x^2-2x-2x+1=0

  2x (2x-1)-(2x-1)=0

  (2x-1)(2x-1)=0

  (2x-1)^2=0

=>2x-1=0 <=> x=1/2

a,  4x^2 - 4x = -1

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

\(\Leftrightarrow\)(2x-1)²              =0 

\(\Leftrightarrow\)2x - 1          = 0 

\(\Leftrightarrow\)x                = 1/2

b, \(\Leftrightarrow\)( 2x + 1)^3 = 0

\(\Leftrightarrow\)2x + 1 = 0 

\(\Leftrightarrow\)x       = -1/2

đúng thì

a) \(4x^2-4x=-1\)

\(\Leftrightarrow4x^2-4x+1=0\)

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

\(\Leftrightarrow2x-1=0\)

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

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

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

\(\Leftrightarrow2x+1=0\)

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

a) (2x)² - 4x = -1

2x - 4x = -1

-2x = -1

x = -1/-2

x = 1/2

d) \(4x^2-9-x\left(2x-3\right)=0\)

\(\Leftrightarrow4x^2-9-2x^2+3x=0\)

\(\Leftrightarrow2x^2+3x-9=0\)

\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)

e) \(x^3+5x^2+9x=-45\)

\(\Leftrightarrow x^3+5x^2+9x+45=0\)

\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)

f) \(x^3-6x^2-x+30=0\)

\(\Leftrightarrow\left(x^3-x^2-6x\right)-\left(5x^2-5x-30\right)=0\)

\(\Leftrightarrow x\left(x^2-x-6\right)-5\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+3x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left[x\left(x-2\right)+3\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow x\in\left\{5;-3;2\right\}\)

a) \(4x^2-4x=-1\)

\(\Leftrightarrow4x\left(x-1\right)=-1\)

\(\Leftrightarrow4x=-1\) hoặc \(x-1=-1\)

\(\Leftrightarrow x=\dfrac{-1}{4}\) hoặc \(x=0\)

Vậy S={\(\dfrac{-1}{4};0\)}

\(\text{a) }4x^2-4x=-1\\ \Leftrightarrow4x^2-4x+1=0\\ \Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\\ \Leftrightarrow\left(2x-1\right)^2=0\\ \Leftrightarrow2x-1=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\\ \text{Vậy }x=\dfrac{1}{2}\\ \)

\(\text{ b) }8x^3+12x^2+6x+1=0\\ \Leftrightarrow\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3=0\\ \Leftrightarrow\left(2x+1\right)^3=0\\ \Leftrightarrow2x+1=0\\ \Leftrightarrow2x=-1\\ \Leftrightarrow x-\dfrac{1}{2}\\ \text{Vậy }x=-\dfrac{1}{2}\)

*) \(4x^2-4x=-1\Leftrightarrow4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\)

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

*)\(8x^3+12x^2+6x+1=0\Leftrightarrow8x^3+8x^2+2x+4x^2+4x+1\)

\(\Leftrightarrow2x\left(4x^2+4x+1\right)+\left(x^2+4x+1\right)=0\Leftrightarrow\left(2x+1\right)\left(4x^2+4x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x+1\right)^2=0\Leftrightarrow\left(2x+1\right)^3=0\Leftrightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)

a,4² - 4x = -1

=> 4x² - 4x + 1 = 0

=> (2x - 1)² = 0

=> 2x -1 = 0

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

b, 8x³ +12x² + 6x +1=0

=> ( 2x +1 )³ = 0

=> 2x + 1 = 0

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

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, x² -  10x = -25                       b, 4x² - 4x = -1                                  c,  8x³ +12x² +6x+1=0

=>x²-10x+25=0                         =>(2x)²-2.2x.1+1=0                            =>(2x+1)³=0

=>(x-5)²=0                               =>(2x-1)²=0                                         =>2x+1=0

=>x-5=0                                    =>2x-1=0                                            =>x = -1/2

=>x=5                                       =>x=1/2

a.x²-10x=-25

x²-10x+25=0

(x-5)²=0

x=5

x² - 10x = -25

=> x² - 10x + 25 = 0

=> (x - 5)² = 0

=> x - 5 = 0

=> x = 5