Ответ:
Пошаговое объяснение:
1)
2)
0\\t^2-7*t-18=0\\D=(-7)^2-4*1*(-18)=49+72=121\\t_1=\frac{-(-7)+\sqrt{121} }{2}=\frac{7+11}{2}=\frac{18}{2}=9\\ t_2=\frac{-(-7)-\sqrt{121} }{2}=\frac{7-11}{2}=\frac{-4}{2} =-2\\ x_{1,2}=б\sqrt{9}=б3" alt="x^4-7*x^2-18=0\\t=x^2, t>0\\t^2-7*t-18=0\\D=(-7)^2-4*1*(-18)=49+72=121\\t_1=\frac{-(-7)+\sqrt{121} }{2}=\frac{7+11}{2}=\frac{18}{2}=9\\ t_2=\frac{-(-7)-\sqrt{121} }{2}=\frac{7-11}{2}=\frac{-4}{2} =-2\\ x_{1,2}=б\sqrt{9}=б3" align="absmiddle" class="latex-formula">
3)
0\\9t^2-19t+2=0\\D=(-19)^2-4*9*2=361-72=289\\t_{1,2}=\frac{-(-19)б\sqrt{289} }{2*9} =\frac{19б17}{18} \\t_1=2, t_2=\frac{1}{9} \\x_{1,2}=б\sqrt{2}\\ x_{3,4}=б\sqrt{\frac{1}{9} }=б\frac{1}{3}\\" alt="9x^4-19x^2+2=0\\t=x^2, t>0\\9t^2-19t+2=0\\D=(-19)^2-4*9*2=361-72=289\\t_{1,2}=\frac{-(-19)б\sqrt{289} }{2*9} =\frac{19б17}{18} \\t_1=2, t_2=\frac{1}{9} \\x_{1,2}=б\sqrt{2}\\ x_{3,4}=б\sqrt{\frac{1}{9} }=б\frac{1}{3}\\" align="absmiddle" class="latex-formula">
4)
0\\5t^2+3t-2=0\\D=3^2-4*5*(-2)=9+80=89\\t_{1,2}=\frac{-3б\sqrt{89} }{5*2} =\frac{-3б\sqrt{89} }{10}\\ t_1=\frac{-3+\sqrt{89} }{10} >0\\x_{1,2}=б\sqrt{\frac{-3+\sqrt{89} }{10} } \\t_2=\frac{-3-\sqrt{89} }{10} <0" alt="5x^4+3x^2-2=0\\t=x^2, t>0\\5t^2+3t-2=0\\D=3^2-4*5*(-2)=9+80=89\\t_{1,2}=\frac{-3б\sqrt{89} }{5*2} =\frac{-3б\sqrt{89} }{10}\\ t_1=\frac{-3+\sqrt{89} }{10} >0\\x_{1,2}=б\sqrt{\frac{-3+\sqrt{89} }{10} } \\t_2=\frac{-3-\sqrt{89} }{10} <0" align="absmiddle" class="latex-formula">