giúp tớ vớiii:((((((

Đáp án:

 `23)`

$(2x+1)^2-4.(x+3)^2=0$

 $⇔ (2x+1)^2-[2.(x+3)]^2=0$

$⇔ (2x+1-2x-6).(2x+1+2x+6)=0$

$⇔ (-5).(4x+7)=0$

$⇔ -20x-35=0$

$⇔ -20x=35$

$⇔ x=-\frac{7}{4}$

$\text{Vậy có 1 giá trị x= $-\dfrac{7}{4}$}$

`=> B`

`24)`

$x^4+3x^3+3x^2+x=0$

$⇔ x.(x^3+1)+3x^2.(x+1)=0$

$⇔ x.(x+1).(x^2-x+1)+3x^2.(x+1)=0$

$⇔ (x+1).(x^3-x^2+x+3x^2)=0$

$⇔ (x+1).(x^3+2x+x)=0$

$⇔ x.(x+1).(x^2+2x+1)=0$

$⇔ x.(x+1).(x+1)^2=0$

$ ⇔x.(x+1)^3=0$

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

$\text{Vậy có 2 giá trị x thảo mãn là x=0 hay x=-1}$

`=>C`