(2x+1)^6 -16(2x+1)^4 = 0

Giải thích các bước giải:

`(2x+1)^6 -16(2x+1)^4 =0`

`<=>(2x+1)^4 -[(2x+1)^2 -16]=0`

`<=>(2x+1)^4 -[(2x+1)^2 -4^2]=0`

`<=>(2x+1)^4 (2x+1-4)(2x+1+4)=0`

`<=>(2x+1)^4 (2x-3)(2x+5)=0`

`<=>`\(\left[ \begin{array}{l}(2x+1)^4=0\\2x-3=0\\2x+5=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}2x+1=0\\2x=3\\2x=-5\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=\dfrac{3}{2}\\x=-\dfrac{5}{2}\end{array} \right.\) 

`\text{Vậy phương trình có tập nghiệm: S}={-1/2 ;3/2 ;-5/2}`

Đáp án:

Cách  `1:`

`( 2x + 1 )^6 - 16 ( 2x + 1 )^4 = 0`

`<=>` `( 2x + 1 )^4 [ ( 2x + 1 )^2 - 16 ] = 0`

`<=>` $\left[\begin{matrix} ( 2x + 1 )^4 = 0\\ ( 2x + 1 )^2 - 16 = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} 2x + 1 = 0\\ ( 2x + 1 )^2 = 16\end{matrix}\right.$

`<=>` $\left[\begin{matrix} 2x = -1\\ ( 2x + 1 )^2 = ( ±4 )^2\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = -\dfrac{1}{2}\\ 2x + 1 = -4\\ 2x + 1 = 4\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = -\dfrac{1}{2}\\ 2x = -5\\ 2x = 3\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = -\dfrac{1}{2}\\ x = -\dfrac{5}{2}\\ x = \dfrac{3}{2}\end{matrix}\right.$

Vậy `:` `S = { -1/2 ; -5/2 ; 3/2 }`.

Cách  `2:`

`( 2x + 1 )^6 - 16 ( 2x + 1 )^4 = 0`

`<=>` `( 2x + 1 )^4 [ ( 2x + 1 )^2 - 16 ] = 0`

`<=>` `( 2x + 1 )^4 [ ( 2x + 1 )^2 - 4^2 ] = 0`

`<=>` `( 2x + 1 )^4 ( 2x + 1 - 4 ) ( 2x + 1 + 4 ) = 0`

`<=>` `( 2x + 1 )^4 ( 2x - 3 ) ( 2x + 5 ) = 0`

`<=>` $\left[\begin{matrix} ( 2x + 1 )^4 = 0\\ 2x - 3 = 0\\ 2x + 5 = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} 2x + 1 = 0\\ 2x = -3\\ 2x = -5\end{matrix}\right.$

`<=>` $\left[\begin{matrix} 2x = -1\\ 2x = -3\\ 2x = -5\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = -\dfrac{1}{2}\\ x = -\dfrac{5}{2}\\ x = \dfrac{3}{2}\end{matrix}\right.$

Vậy `:` `S = { -1/2 ; -5/2 ; 3/2 }`.