Gọi \(z_1,z_2,z_3,z_4\) lần lượt là các nghiệm của phương trình \(8z^4+8z^3=z+1\). Giá trị của \(\left|z_1\right|.\left|z_2\right|.\left|z_3\right|.\left|z_4\right|\) bằng
\(\dfrac{1}{2}\).\(\dfrac{1}{4}\).\(\dfrac{1}{8}\).\(\dfrac{1}{16}\).Hướng dẫn giải:\(8z^4+8z^3=z+1\Leftrightarrow8z^3\left(z+1\right)-\left(z+1\right)=0\)
\(\Leftrightarrow\left(z+1\right)\left[8z^3-1\right]=0\)
\(\Leftrightarrow\left(z+1\right)\left(2z-1\right)\left(4z^2+2z+1\right)=0\)
\(\Leftrightarrow\left(z+1\right)\left(2z-1\right)\left[z^2+2z+1+3z^2\right]=0\)
\(\Leftrightarrow\left(z+1\right)\left(2z-1\right)\left[\left(z+1\right)^2-\left(\sqrt{3}zi\right)^2\right]=0\)
\(\Leftrightarrow\left(z+1\right)\left(2z-1\right)\left(z+1-\sqrt{3}zi\right)\left(z+1+\sqrt{3}zi\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}z+1=0\\2z-1=0\\z+1-\sqrt{3}zi=0\\z+1+\sqrt{3}zi=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}z=-1\\z=\frac{1}{2}\\z=\frac{-1}{1-\sqrt{3}i}\\z=\frac{-1}{1+\sqrt{3}i}\end{array}\right.\)
Vậy \(\left|z_1\right|.\left|z_2\right|.\left|z_3\right|.\left|z_4\right|=\left|-1.\right|.\left|\frac{1}{2}\right|.\left|\frac{-1}{1-\sqrt{3}i}\right|.\left|\frac{-1}{1+\sqrt{3}i}\right|\)
\(=\left|-1.\frac{1}{2}.\frac{-1}{1-\sqrt{3}i}.\frac{-1}{1+\sqrt{3}i}\right|\)
\(=\left|-\frac{1}{2}.\frac{1}{1-3i^2}\right|=\frac{1}{8}\).