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

Lời giải:

PT $\Leftrightarrow 8x^3-16x^2+6x+2=0$

$\Leftrightarrow (8x^3-8x^2)-(8x^2-8x)-(2x-2)=0$

$\Leftrightarrow 8x^2(x-1)-8x(x-1)-2(x-1)=0$

$\Leftrightarrow (x-1)(8x^2-8x-2)=0$

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

$\Leftrightarrow x-1=0$ hoặc $4x^2-4x-1=0$

Nếu $x-1=0\Leftrightarrow x=1$ 

Nếu $4x^2-4x-1=0$

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

$\Leftrightarrow (2x-1-\sqrt{2})(2x-1+\sqrt{2})=0$

$\Leftrightarrow x=\frac{1\pm \sqrt{2}}{2}$

( 4x - 1 )³ + ( 3 - 4x )( 9 + 12x + 16x² ) = ( 8x - 1 )( 8x + 1 ) - ( 3x - 5 )

<=> 64x³ - 48x² + 12x - 1 + [ 3³ - ( 4x )³ ] = ( 8x )² - 1 - 3x + 5

<=> 64x³ - 48x² + 12x - 1 + 27 - 64x³ = 64x² - 3x + 4

<=> -48x² + 12x + 26 = 64x² - 3x + 4

<=> -48x² + 12x + 26 - 64x² + 3x - 4 = 0

<=> -112x² + 15x + 22 = 0 (*)

\(\Delta=b^2-4ac=15^2-4\cdot\left(-112\right)\cdot22=225+9856=10081\)

\(\Delta>0\)nên (*) có hai nghiệm phân biệt

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{\sqrt{10081}-15}{-224}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-15-\sqrt{10081}}{-224}\end{cases}}\)

Lớp 8 sao nghiệm xấu thế -..-

Dạ em cảm ơn

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

( 4x - 1 )³ + ( 3 - 4x )( 9 + 12x + 16x² ) = ( 8x - 1 )( 8x + 1 ) - ( 3x - 5 )

<=> 64x³ - 48x² + 12x - 1 + [  3³ - ( 4x )³ ] = ( 8x )² - 1² - 3x + 5

<=> 64x³ - 48x² + 12x - 1 + 27 - 64x³ = 64x² - 1 - 3x + 5

<=> 64x³ - 48x² + 12x - 64x³ - 64x² + 3x = -1 + 5 + 1 - 27

<=> -112x² + 15x = -22

<=> -112x² + 15x + 22 = 0 (*) ( lại phải xài Delta :(( )

\(\Delta=b^2-4ac=15^2-4\cdot\left(-112\right)\cdot22=225+9856=10081\)

\(\Delta>0\)nên (*) có hai nghiệm phân biệt 

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-15+\sqrt{10081}}{-224}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-15-\sqrt{10081}}{-224}\end{cases}}\)

Nghiệm xấu quá -..-

a: ĐKXD: x<>0

\(\dfrac{14x^3+12x^2-14x}{2x}=\left(x+2\right)\left(3x-4\right)\)

=>\(\dfrac{2x\left(7x^2+6x-7\right)}{2x}=\left(x+2\right)\left(3x-4\right)\)

=>\(7x^2+6x-7=3x^2-4x+6x-8\)

=>\(7x^2+6x-7=3x^2+2x-8\)

=>\(4x^2+4x+1=0\)

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

=>2x+1=0

=>x=-1/2(nhận)

b: \(\left(4x-5\right)\left(6x+1\right)-\left(8x+3\right)\left(3x-4\right)=15\)

=>\(24x^2+4x-30x-5-\left(24x^2-32x+9x-12\right)=15\)

=>\(24x^2-26x-5-24x^2+23x+12=15\)

=>-3x+7=15

=>-3x=8

=>\(x=-\dfrac{8}{3}\)

Bài làm:

Ta có: \(\left(4x-1\right)^3+\left(3-4x\right)\left(9+12x+16x^2\right)=\left(8x-1\right)\left(8x+1\right)-\left(3x-5\right)\)

\(\Leftrightarrow64x^3-48x^2+12x-1+27-64x^3-64x^2+1+3x-5=0\)

\(\Leftrightarrow15x+22=0\)

\(\Leftrightarrow15x=-22\)

\(\Rightarrow x=-\frac{22}{15}\)

bạn ơi cho mình hỏi -48x²-64x² đâu ạ