$\text{Đáp án + giải thích các bước giải}$

$\text{a) $\dfrac{64x^{3} + 1 }{16x^{2} - 1}$}$ $\text{=$\dfrac{A}{4x - 1}$}$

$\text{⇒ A. ( 16x² - 1 ) = ( 64x³ + 1 ) ( 4x - 1 )}$

$\text{⇒ A ( 4x - 1 ) ( 4x + 1 ) = ( 4x + 1 ) ( 16x² - 4x + 1 ) ( 4x - 1 ) }$

$\text{⇒ A ( 16x² - 1 ) = ( 16x² - 1 ) ( 16x² - 4x + 1 )}$

$\text{⇔ A = $\dfrac{( 16x² - 1 ) ( 16x² - 4x + 1 )}{16x^{2} - 1}$}$

$\text{⇔ A = 16x² - 4x + 1}$

$\text{b) $\dfrac{2x - y - 1}{4x - 2y}$ }$ $\text{= $\dfrac{4x^{2} - 2x - y^{2} - y}{B}$}$

$\text{⇒ D(2x-y-1)=(4x-2y)(4x²-2x-y²-y)}$

$\text{⇒ D ( 2x - y - 1 ) = ( 4x - 2y ) ( 4x² - y² - 2x - y )}$

$\text{⇒ D ( 2x - y - 1 ) = ( 4x - 2y ) [ ( 2x - y ) ( 2x + y ) - ( 2x + y ) ]}$

$\text{⇒ D ( 2x - y - 1 ) = ( 4x - 2y ) ( 2x + y ) ( 2x - y - 1 ) }$

$\text{⇔ D = $\dfrac{( 4x - 2y ) ( 2x + y ) ( 2x - y - 1 )}{2x - y - 1}$ }$

$\text{⇔ D = ( 4x - 2y ) ( 2x + y )}$

$\text{⇔ D = 8x² - 4y²}$

$\text{Cho mình xin câu tl hay nhất nha}$

Đáp án:

 a)

⇔A.(16x²-1)=(64x³+1)(4x-1)

⇔A(4x-1)(4x+1)=(4x+1)(16x²-4x+1)(4x-1)

⇔A(16x²-1)=(16x²-1)(16x²-4x+1)

⇔A=$\frac{(16x^{2}-1)(16x^{2}-4x+1)}{16x^{2}-1}$ 

⇔A=16x²-4x+1

b)

⇔D(2x-y-1)=(4x-2y)(4x²-2x-y²-y)

⇔D(2x-y-1)=(4x-2y)(4x²-y²-2x-y)

⇔D(2x-y-1)=(4x-2y)[ (2x-y)(2x+y)-(2x+y)]

⇔D(2x-y-1)=(4x-2y)(2x+y)(2x-y-1)

⇔D=$\frac{(4x-2y)(2x+y)(2x-y-1)}{(2x-y-1)}$ 

⇔D=(4x-2y)(2x+y)

       =8x²-4y²