(2x+1)(x+1)^2(2x+3)=18

$(2x+1)(x+1)^2(2x+3)=18\\⇔[(2x+1)(2x+3)](x^2+2x+1)=18\\⇔(4x^2+8x+3)(x^2+2x+1)=18$

Đặt $x^2+2x+1=y(y\ge 0)$

Phương trình trở thành $(4y-1)y=18$

$⇔4y^2-y-18=0\\⇔4y^2-9y+8y-18=0\\⇔y(4y-9)+2(4y-9)=0\\⇔(y+2)(4y-9)=0\\⇔\left[\begin{array}{1}y+2=0\\4y-9=0\end{array}\right.\\⇔\left[\begin{array}{1}y=-2\\4y=9\end{array}\right.\\⇔\left[\begin{array}{1}y=-2(KTM)\\y=\dfrac{9}{4}(TM)\end{array}\right.$

Với $y=\dfrac{9}{4}⇒x^2+2x+1=\dfrac{9}{4}$

$⇔4x^2+8x+4=9\\⇔4x^2+8x-5=0\\⇔4x^2+10x-2x-5=0\\⇔2x(2x+5)-(2x+5)=0\\⇔(2x-1)(2x+5)=0\\⇔\left[\begin{array}{1}2x-1=0\\2x+5=0\end{array}\right.\\⇔\left[\begin{array}{1}2x=1\\2x=-5\end{array}\right.\\⇔\left[\begin{array}{1}x=\dfrac{1}{2}\\x=-\dfrac{5}{2}\end{array}\right.$

Vậy phương trình có tập nghiệm $S=\left\{\dfrac{1}{2};-\dfrac{5}{2}\right\}$

Đáp án + Giải thích các bước giải:

`( 2x + 1 ) ( x + 1 )^2 ( 2x + 3 ) = 18`

`= ( 2x + 1 ) ( 2x + 3 ) ( x + 1 )^2 - 18`

`= ( 4x^2 + 6x + 2x + 3 ) ( x^2 + 2x + 1 ) - 18 `

`= 4x^4 + 8x^3 + 12x^2 + 8x^3 + 16x^2 + 21x - 5x^2 - 10x - 15 `

`= ( 4x^2 + 8x^3 + 12x^2 ) + ( 8x^3 + 16x^2 + 21x ) - ( 5x^2 - 10x - 15 ) `

`= 4x^2 ( x^2 + 2x + 3 ) + 8x ( x^2 + 2x + 3 ) - 5 ( x^2 + 2x + 3 ) `

`= ( x^2 + 2x + 3 ) ( 4x^2 + 8x - 5 ) `

`= ( x^2 + 2x + 3 ) ( 4x^2 + 10x - 2x - 5 ) `

`= ( x^2 + 2x + 3 ) [ ( 4x^2 + 10x ) - ( 2x + 5 ) ]`

`= ( x^2 + 2x + 3 ) [ 2x ( x + 5 ) - ( 2x + 5 ) ]`

`= ( x^2 + 2x + 3 ) ( 2x - 1 ) ( 2x + 5 ) `