Question

Đề : Giải phương trình (bằng công thức nghiệm)

Answer 1

d, \(\Delta=\left[-\left(\sqrt{5}-3\right)\right]^2-4.1.\left(-3\sqrt{5}\right)=5-6\sqrt{5}+9+12\sqrt{5}=14+6\sqrt{5}\)

\(x_1=\dfrac{-\left[-\left(\sqrt{5}-3\right)\right]+\sqrt{14+6\sqrt{5}}}{2}=\dfrac{\sqrt{5}-3+\sqrt{14+2\sqrt{45}}}{2}=\dfrac{\sqrt{5}-3+\sqrt{9+2\sqrt{9}\sqrt{5}+5}}{2}=\dfrac{\sqrt{5}-3+\sqrt{\left(3+\sqrt{5}\right)^2}}{2}=\dfrac{\sqrt{5}-3+3+\sqrt{5}}{2}=\dfrac{2\sqrt{5}}{2}=\sqrt{5}\)

\(x_2=\dfrac{-\left[-\left(\sqrt{5}-3\right)\right]-\sqrt{14+6\sqrt{5}}}{2}=\dfrac{\sqrt{5}-3-\sqrt{14+2\sqrt{45}}}{2}=\dfrac{\sqrt{5}-3-\sqrt{9+2\sqrt{9}\sqrt{5}+5}}{2}=\dfrac{\sqrt{5}-3-\sqrt{\left(3+\sqrt{5}\right)^2}}{2}=\dfrac{\sqrt{5}-3-3-\sqrt{5}}{2}=\dfrac{-6}{2}=-3\)

Answer 2

\(e,\Delta=\left[-\left(\sqrt{7}+\sqrt{3}\right)\right]^2-4.1.\sqrt{21}=7+2\sqrt{21}+3-4\sqrt{21}=10-2\sqrt{21}\)

\(x_1=\dfrac{-\left[-\left(\sqrt{7}+\sqrt{3}\right)\right]+\sqrt{10-2\sqrt{21}}}{2}=\dfrac{\sqrt{7}+\sqrt{3}+\sqrt{7-2\sqrt{7}\sqrt{3}+3}}{2}=\dfrac{\sqrt{7}+\sqrt{3}+\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}{2}=\dfrac{\sqrt{7}+\sqrt{3}+\sqrt{7}-\sqrt{3}}{2}=\dfrac{2\sqrt{7}}{2}=\sqrt{7}\)

\(x_1=\dfrac{-\left[-\left(\sqrt{7}+\sqrt{3}\right)\right]-\sqrt{10-2\sqrt{21}}}{2}=\dfrac{\sqrt{7}+\sqrt{3}-\sqrt{7-2\sqrt{7}\sqrt{3}+3}}{2}=\dfrac{\sqrt{7}+\sqrt{3}-\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}}{2}=\dfrac{\sqrt{7}+\sqrt{3}-\sqrt{7}+\sqrt{3}}{2}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)