27x^3+27x^2+9x+1-(1+3x)^2=0 tìm x mong các anh chị giúp giùm e ạ 04/07/2021 Bởi Bella 27x^3+27x^2+9x+1-(1+3x)^2=0 tìm x mong các anh chị giúp giùm e ạ
`⇒27x^3+27x^2+9x+1-(1+3x)^2=0` `⇒27x^3+27x^2+9x+1-1+6x-9x^2=0` `⇒27x^3+18x^2+3x=0` `⇒3x(9x^2+6x+1)=0` $⇒\left[ \begin{array}{l}3x=0\\9x^2+6x+5=0\end{array} \right.$ $⇒\left[ \begin{array}{l}x=0\\(3x+1)^2=0\end{array} \right.$ $⇒\left[ \begin{array}{l}x=0\\x=-\frac{1}{3}\end{array} \right.$ Vậy `x=0; x=-1/3` Bình luận
Đáp án: Giải thích các bước giải: `27x^3+27x^2+9x+1-(1+3x)^2=0` `<=>(3x+1)^3-(3x+1)^2=0` `<=>(3x+1)^2(3x+1-1)=0` `<=>3x(3x+1)=0` `=>`\(\left[ \begin{array}{l}3x=0\\3x+1=0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=0\\x=-\dfrac{1}{3}\end{array} \right.\) Bình luận
`⇒27x^3+27x^2+9x+1-(1+3x)^2=0`
`⇒27x^3+27x^2+9x+1-1+6x-9x^2=0`
`⇒27x^3+18x^2+3x=0`
`⇒3x(9x^2+6x+1)=0`
$⇒\left[ \begin{array}{l}3x=0\\9x^2+6x+5=0\end{array} \right.$
$⇒\left[ \begin{array}{l}x=0\\(3x+1)^2=0\end{array} \right.$
$⇒\left[ \begin{array}{l}x=0\\x=-\frac{1}{3}\end{array} \right.$
Vậy `x=0; x=-1/3`
Đáp án:
Giải thích các bước giải:
`27x^3+27x^2+9x+1-(1+3x)^2=0`
`<=>(3x+1)^3-(3x+1)^2=0`
`<=>(3x+1)^2(3x+1-1)=0`
`<=>3x(3x+1)=0`
`=>`\(\left[ \begin{array}{l}3x=0\\3x+1=0\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=0\\x=-\dfrac{1}{3}\end{array} \right.\)