1 câu trả lời
$(x^2-1)(x+3)(x+5)-9=0$
$⇔(x-1)(x+1)(x+3)(x+5)-9=0$
$⇔(x-1)(x+5)(x+1)(x+3)-9=0$
$⇔(x^2+4x-5)(x^2+4x+3)-9=0$
Đăt: $x^2+4x=a$
$⇔(a-5)(a+3)-9=0$
$⇔a^2-2a-15-9=0$
$⇔a^2-2a-24=0$
$⇔a^2+4a-6a-24=0$
$⇔(a+4)(a-6)=0$
$⇔(x^2+4x+4)(x^2+4x-6)=0$
$⇔(x+2)^2(x^2+4x-6)=0$
$⇔$ \(\left[ \begin{array}{l}x+2=0\\x^2+4x-6=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=-2\\x^2+4x-6=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-2\\(x+2)^2=10\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-2\\x=±\sqrt{10}-2\end{array} \right.\)
Vậy `S={-2;±\sqrt{10}-2}`
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