Question

Giúp mik câu 3 d vs ạ

Answer 1

Câu 3:

3)

d) \(3x^2+x-6-\sqrt{2}=0\)

\(\Leftrightarrow x^2+\dfrac{1}{3}x-2-\dfrac{\sqrt{2}}{3}=0\)

\(\Leftrightarrow x^2+\dfrac{1}{3}x-\dfrac{6+\sqrt{2}}{3}=0\)

\(\Leftrightarrow x^2+2.\dfrac{1}{6}x+\dfrac{1}{36}-\dfrac{6+\sqrt{2}}{3}-\dfrac{1}{36}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{6}\right)^2-\dfrac{73+12\sqrt{2}}{36}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{6}+\dfrac{\sqrt{73+12\sqrt{2}}}{6}\right)\left(x+\dfrac{1}{6}-\dfrac{\sqrt{73+12\sqrt{2}}}{6}\right)=0\)

\(\Leftrightarrow x=\dfrac{-1\pm\sqrt{73+12\sqrt{2}}}{6}\)

-Vậy \(S=\left\{\dfrac{-1\pm\sqrt{73+12\sqrt{2}}}{6}\right\}\)