( 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 -..-

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

2a) pt <=> (x + 6)^2 = 0

<=> x = -6

b) pt <=> (4x - 1)^2 = 0

<=> x = 1/4

c) pt<=> (x + 1)^3 = 0

<=> x = -1

Bài 1:

a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)

\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)

\(=32x^2+18y^2\)

b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)

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

\(=-12x^2-24\)

Bài 2: 

a: Ta có: \(x^2+12x+36=0\)

\(\Leftrightarrow x+6=0\)

hay x=-6

b: Ta có: \(16x^2-8x+1=0\)

\(\Leftrightarrow4x-1=0\)

hay \(x=\dfrac{1}{4}\)

Bài 1: 

a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)

\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)

\(=32x^2+18y^2\)

b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)

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

\(=-12x^2-24\)

c: Ta có: \(C=\left(x+2y\right)^2+2\left(x+2y\right)\left(x-2y\right)+\left(x-2y\right)^2\)

\(=\left(x+2y+x-2y\right)^2\)

\(=4x^2\)

\(4x\left(12x-9\right)-8x\left(6x-5\right)=1\)

\(\Leftrightarrow48x^2-36x-48x^2+40x=1\)

\(\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\)

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

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