Câu hỏi của roseandlisa

Question

1.Giải phương trình:

3x^4+x^2-4=03$x^4$ +<...

Minh Nhân · Accepted Answer

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

$\Leftrightarrow3x^4-3x^2+4x^2-4=0$

$\Leftrightarrow3x^2\cdot\left(x^2-1\right)+4\cdot\left(x^2-1\right)=0$

$\Leftrightarrow\left(x^2-1\right)\left(3x^2+4\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\3x^2+4=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\pm1\x^2=-\dfrac{4}{3}\left(l\right)\end{matrix}\right.$

$S=\left\{\pm1\right\}$

Câu trả lời:

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

$\Leftrightarrow3x^4-3x^2+4x^2-4=0$

$\Leftrightarrow3x^2\cdot\left(x^2-1\right)+4\cdot\left(x^2-1\right)=0$

$\Leftrightarrow\left(x^2-1\right)\left(3x^2+4\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\3x^2+4=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\pm1\x^2=-\dfrac{4}{3}\left(l\right)\end{matrix}\right.$

$S=\left\{\pm1\right\}$

Trần Ái Linh · Answer

Đặt `x^2=t(t>=0)`

Ta có PT: `3t^2+t-4=0`

`3+1-4=0`

`=> t_1 = 1 ; t_2 = -4/3 (L)`

`=> x^2=1`

`<=> x=\pm 1`

Vậy `S={\pm 1}`.