ОДЗ: x≥0
пусть √x=t, t≥0, тогда:
t²+t-13=0
D= 1+52= 53
t1= (-1-√53)/2
t2= (-1+√53)/2
√x= (-1+√53)/2
x= (53-2√53+1)/4= (54-2√53)/4
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Verified answer
ОДЗ: x≥0
пусть √x=t, t≥0, тогда:
t²+t-13=0
D= 1+52= 53
t1= (-1-√53)/2
t2= (-1+√53)/2
√x= (-1+√53)/2
x= (53-2√53+1)/4= (54-2√53)/4
Verified answer