Question

Решите уравнения: а)4х^4-5×^2+1=0
б)(2×^2-3)^2-4(2×^2-3)=21 d) (x^2-2×)^2-3=2(×^2-2×)

Answer 1

a)x^{2} =t 4 t^{2} -5t+1=0 D=25-4*4*1=9= 3^{2} t1=(5+3)8=8/8=1 t2=(5-3)/8=2/8=1/4=0.25 x^{2} =1 x=1 и x=-1 x^{2} =0.25 x=0.5 и x=-0.5 

б)t= (2x^{2} -3) t^{2}-4t-21=0 D=16+84=100= 10^{2} t1=(4+10)/2=14/2=7 t2=(4-10)/2=6-/2=-3 2 x^{2} -3=7
x= sqrt{5} и x=- sqrt{5} 2 x^{2} -3=-3 x=0

d)x^{2} -2=t   t^{2}-2t-3=0 D=16= 4^{2} t1=(2+4)/2=6/2=3 t2=(2-4)/2=-2/2=-1 x^{2} -2x=3 x^{2} -2x-3=0 D=16= 4^{2} x1=6/2=3 x2=-2/2=-1 x^{2} -2x=-1 D=0 x=-2/2=-1