Sợ mấy cái bài kiểu này luôn:(
\(x^3+2x^2-4x+\dfrac{8}{3}=0\) (1)
Đặt t = \(x+\dfrac{2}{3}\Rightarrow x=t-\dfrac{2}{3}\)
KHi đó \(\left(1\right)\Leftrightarrow\left(t-\dfrac{2}{3}\right)^3+2\left(t-\dfrac{2}{3}\right)^2-4\left(t-\dfrac{2}{3}\right)+\dfrac{8}{3}=0\)\(t^3-\dfrac{16}{3}t+\dfrac{160}{27}=0\) (2)
Đặt \(y=\dfrac{t}{\dfrac{8}{3}}\Rightarrow t=\dfrac{8}{3}y\)
\(\Rightarrow\left(2\right)\Leftrightarrow4y^3-3y=-\dfrac{5}{32}\)
Đặt \(a=\sqrt[3]{-\dfrac{5}{32}+\sqrt{-\dfrac{5}{32}^2+1}}\) và \(\alpha=\dfrac{1}{2}\left(a-\dfrac{1}{a}\right)\) ta được:
\(4\alpha^3-3\alpha=-\dfrac{5}{32}\)
\(\Rightarrow y=\dfrac{1}{2}\left(\sqrt[3]{-\dfrac{5}{32}+\sqrt{-\dfrac{5}{32}^2+1}}+\sqrt[3]{-\dfrac{5}{32}-\sqrt{-\dfrac{5}{32}^2+1}}\right)\)=-0,06155982175
\(\Rightarrow t=\dfrac{8}{3}y=-0,1641595247\)
\(\Rightarrow x=t-\dfrac{2}{3}=-0,8308261913\)
mk có cách giải ngắn hơn bạn dưới
x^3=-2x^2+4x-8/3
3x^2=-8-6x^2+12x
3x^3 +x^3=x^3-3x^2+3x.4-2^3
4x^3=(x-2)^3
x= \(\dfrac{-2}{\sqrt[3]{4}-1}\)
