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danilcapitan
@danilcapitan
July 2022
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Решите систему уравнений:
[tex] \left \{ {{2x+y=1} \atop {3x^2-xy=18}} \right. [/tex]
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армения2017
У=1-2х
3х^2-х+2х^2=18
5х^2-х-18=0
D=1+4•5•18=361>0
x1=(1+√361)/10=(1+19)/10=2
x2=(1-19)/10=-18/10=-9/5
y1=1-2x1=1-4=-3
y2=1-2x2=1-2(-9/5)=1+18/5=23/5
ответ (2;-3) (-9/5;23/5)
2 votes
Thanks 1
danilcapitan
Спасибо :)
danilcapitan
Правда я уже решил, перепроверь себя, там должно быть (-1.8 4.6) (2 -3 )
армения2017
-1,8=-18/10=-9/5 4,6=46/10=23/5
danilcapitan
А, прости.
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Answers & Comments
3х^2-х+2х^2=18
5х^2-х-18=0
D=1+4•5•18=361>0
x1=(1+√361)/10=(1+19)/10=2
x2=(1-19)/10=-18/10=-9/5
y1=1-2x1=1-4=-3
y2=1-2x2=1-2(-9/5)=1+18/5=23/5
ответ (2;-3) (-9/5;23/5)