giải phương trình sau:
[TEX](z+1)^4 +2*(z+1)^2 +(z+4)^2+1=0[/TEX]
[TEX]\Leftrightarrow (z+1)^4+2(z+1)^2+1=-(z+4)^2[/TEX]
[TEX]\leftrightarrow \[ (z+1\)^2+1\]^2=i^2(z+4)^2[/TEX]
[TEX]\leftrightarrow \left[ (z+1\)^2+1=i \(z+4\)\ \ \(1\) \\(z+4\)^2+1=-i\(z+4\) \ \ \(2\)[/TEX]
[TEX]\(1\)\leftrightarrow z^2+\(2-i\)z-4i+2=0[/TEX]
[TEX]\Delta:=11+12i=\(a+bi\)^2[/TEX]
[TEX]\leftrightarrow\left{a^2=\frac{\sqrt{265}+11}{2}\\ b^2=\frac{\sqrt{265}-11}{2}\\2ab>0[/TEX]
Do đó ta có
[TEX]\Delta:= \(\sqrt{\frac{\sqrt{265}+11}{2}}+\sqrt{\frac{\sqrt{265}-11}{2}}i\)^2[/TEX]
[TEX]\(1\)\leftrightarrow\left[ z_1:=\frac{\sqrt{\frac{\sqrt{265}+11}{2}}-2+\(1+\sqrt{\frac{\sqrt{265}-11}{2}}\)i}{2}\\z_2:= \frac{-\sqrt{\frac{\sqrt{265}+11}{2}}-2+\(1-\sqrt{\frac{\sqrt{265}-11}{2}}\)i}{2}\ [/TEX]
Do đó theo định lí [TEX]Gause[/TEX] thì ta có
[TEX](2)\leftrightarrow\left[ z_3:=\frac{\sqrt{\frac{\sqrt{265}+11}{2}}-2-\(1+\sqrt{\frac{\sqrt{265}-11}{2}}\)i}{2}\\z_4:= \frac{-\sqrt{\frac{\sqrt{265}+11}{2}}-2-\(1-\sqrt{\frac{\sqrt{265}-11}{2}}\)i}{2}\ [/TEX]
Vậy phương trình có bốn nghiệm [TEX]z_1;z_2;z_3;z_4[/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn