Toán Lớp 10: Giúp với . Giải pt can2x^2-1=x-3
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Toán Lớp 10: Giúp với . Giải pt can2x^2-1=x-3
Comments ( 2 )
$\sqrt{2x^2-1}=x-3$ $;$ $(ĐK:x\geq 3)$
$⇔2x^2-1=x^2-6x+9$
$⇔x^2-6x+9=2x^2-1$
$⇔x^2-6x+9+1=2x^2-1+1$
$⇔x^2-6x+10=2x^2$
$⇔x^2-6x+10-2x^2=2x^2-2x^2$
$⇔-x^2-6x+10=0$
\(⇔\left[ \begin{array}{l}x=\dfrac{-\left(-6\right)+2\sqrt{19}}{2\left(-1\right)}\\x=\dfrac{-\left(-6\right)-2\sqrt{19}}{2\left(-1\right)}\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=-3-\sqrt{19}(tm)\\x=\sqrt{19}-3(ktm)\end{array} \right.\)
$\sqrt{2x^2-1}$ =x-3 ( Đk: x\ge3)
<=>2x^2-1=(x-3)^2
<=>2x^2-1=x^2-6x+9
<=>x^2+6x-10=0
<=>\(\left[ \begin{array}{l}x=-3+\sqrt{19}(TM)\\x=-3-\sqrt{19}(loại)\end{array} \right.\)