Toán Lớp 8: Giải phương trình :
$3x^{2}$ + $x^{}$ – $6^{}$ – $\sqrt[]{2}$ = $0^{}$
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Toán Lớp 8: Giải phương trình :
$3x^{2}$ + $x^{}$ – $6^{}$ – $\sqrt[]{2}$ = $0^{}$
Comments ( 2 )
3 x^2 + x – 6 – sqrt(2) = 0
⇔3 x^2 + x = 6 + sqrt(2)
⇔3 (x + 1/6)^2 + 1/12 (12 (-6 – sqrt(2)) – 1) = 0
$⇔3\left(x+\dfrac{1}{6}\right)^2+\dfrac{1}{12}\left(12\left(-6-\sqrt{2}\right)-1\right)-\dfrac{1}{12}\left(12\left(-6-\sqrt{2}\right)-1\right)=0-\dfrac{1}{12}\left(12\left(-6-\sqrt{2}\right)-1\right)$
$⇔3\left(x+\dfrac{1}{6}\right)^2=-\dfrac{12\left(-6-\sqrt{2}\right)-1}{12}$
$⇔\dfrac{3\left(x+\dfrac{1}{6}\right)^2}{3}=\dfrac{-\dfrac{12\left(-6-\sqrt{2}\right)-1}{12}}{3}$
$⇔\left(x+\dfrac{1}{6}\right)^2=-\dfrac{-73-12\sqrt{2}}{36}$
\(⇔\left[ \begin{array}{l}\left(x+\dfrac{1}{6}\right)^2=\sqrt{-\dfrac{-73-12\sqrt{2}}{36}}\\\left(x+\dfrac{1}{6}\right)^2=-\sqrt{-\dfrac{-73-12\sqrt{2}}{36}}\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=\sqrt{2}\\x=-\dfrac{3\sqrt{2}+1}{3}\end{array} \right.\)
$\text{Vậy pt có nghiệm {$\sqrt{2}$ ; $-\dfrac{3\sqrt{2}+1}{3}$}}$
Giải đáp:
$x=\sqrt{2}$ hoăc $x=-\dfrac{3\sqrt{2}+1}{3}$
Lời giải và giải thích chi tiết:
$3x^2+x-6-\sqrt{2}=0\\⇔3x^2-3\sqrt{2}x+3\sqrt{2}x+x-6-\sqrt{2}=0\\⇔(3x^2-3\sqrt{2}x)+[(3\sqrt{2}+1)x-(6+\sqrt{2})]=0\\⇔3x(x-\sqrt{2})+[(3\sqrt{2}+1)x-\sqrt{2}(3\sqrt{2}+1)]=0\\⇔3x(x-\sqrt{2})+(3\sqrt{2}+1)(x-\sqrt{2})=0\\⇔(x-\sqrt{2})[3x+(3\sqrt{2}+1)]=0\\⇔\left[\begin{array}{l}x-\sqrt{2}=0\\3x+3\sqrt{2}+1=0\end{array}\right.\\⇔\left[\begin{array}{l}x=\sqrt{2}\\x=-\dfrac{3\sqrt{2}+1}{3}\end{array}\right.$
Vậy $x=\sqrt{2}$ hoăc $x=-\dfrac{3\sqrt{2}+1}{3}$