Question

giải phương trình:

2a³-2a²+1=0

Answer 1

2a^3-2a^2+1=0

Δ=b^2-3ac=(-2)^2-3*2*0=4

\(k=\dfrac{9abc-2b^3-27a^2d}{2\cdot\sqrt{4^3}}=\dfrac{9\cdot2\cdot\left(-2\right)\cdot0-2\cdot0^3-27\cdot2^2\cdot1}{2\cdot8}=\dfrac{-108}{16}=\dfrac{-27}{4}\)

=>PT có nghiệm duy nhất là:

\(x=\dfrac{\sqrt{\dfrac{27}{4}}}{3\cdot2}\left(\sqrt[3]{-\dfrac{27}{4}+\sqrt{\left(-\dfrac{27}{4}\right)^2+1}}+\sqrt[3]{-\dfrac{27}{4}-\sqrt{\left(-\dfrac{27}{4}\right)^2+1}}\right)-\dfrac{-2}{3\cdot2}\)

\(=\dfrac{3\sqrt{3}}{12}\cdot\left(\sqrt[3]{-\dfrac{27}{4}+\dfrac{\sqrt{745}}{4}}+\sqrt[3]{-\dfrac{27}{4}-\dfrac{\sqrt{745}}{4}}\right)+\dfrac{1}{3}\)

\(\simeq-0.57\)