Question

Lập phương trình bậc hai có hai nghiệm là : \(\dfrac{2}{2-\sqrt{2}}\)và\(\dfrac{2}{2+\sqrt{2}}\)

Answer 1

Ta có x₁=\(\dfrac{2}{2-\sqrt{2}},x_2=\dfrac{2}{2+\sqrt{2}}\)

=> x₁+x₂=\(\dfrac{2}{2-\sqrt{2}}+\dfrac{2}{2+\sqrt{2}}=4\)

x₁x₂=\(\dfrac{2}{2-\sqrt{2}}.\dfrac{2}{2+\sqrt{2}}=2\)

+=> ct : x²-4x+2=0

Answer 2

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{2}{2-\sqrt{2}}=\dfrac{2.\left(2+\sqrt{2}\right)}{4-2}=2+\sqrt{2}\\x_2=\dfrac{2}{2+\sqrt{2}}=\dfrac{2.\left(2-\sqrt{2}\right)}{4-2}=2-\sqrt{2}\end{matrix}\right.\)

pt bậc 2 có \(\left\{{}\begin{matrix}\Delta'=\sqrt{2}\\b'=-2\end{matrix}\right.\) \(\Leftrightarrow x^2-4x+2=0\)