[tex]a)x^4+9=6x^2\\ x^4-6x^2+9=0\\(x^2-3)^2=0\\|x^2-3|=0\\x^2-3=0\\x^2=3\\|x|=\sqrt{3} \\x_1=\sqrt{3} \\x_2=-\sqrt{3} \\\\b)x^2-1=x^4-x^2\\x^2-1=x^2(x^2-1)\\x^2-1-x^2(x^2-1)=0\\(x^2-1)(1-x^2)=0\\(x-1)(x+1)(1-x)(1+x)=0\\x_1=1\\x_2=-1\\\\c)16x^4-1=8x^2\\16x^4-8x^2-1=0\\x^2=t\\t\geq 0\\16t^2-8t-1=0[/tex]
∆=[tex]64+64=128[/tex]
[tex]\sqrt{128}=8\sqrt{2} \\t_1=\frac{8+8\sqrt{2} }{32} =\frac{1+\sqrt{2} }{4}[tex]x^2=\frac{1+\sqrt{2} }{4}\\t_2=\frac{1-\sqrt{2} }{4}[/tex]
t≥0 od zera, więc bierzemy tylko [tex]t_1[/tex]
[tex]|x|=\frac{\sqrt{1+\sqrt{2} } }{2} \\x_1=\frac{\sqrt{1+\sqrt{2} } }{2} \\x_2=-\frac{\sqrt{1+\sqrt{2} } }{2}[/tex]
[tex]d)12x^2+6=2-9x^4\\9x^4+12x^2+4=0\\(3x^2+2)^2=0\\|3x^2+2|=0\\[/tex]
Nigdy. Równanie nie ma rozwiązania. x∈∅
e) tak jak przykład c
