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Бекш1
@Бекш1
August 2022
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128+2x^4=40x^2 как решить?
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IrkaShevko
Verified answer
X² = t ≥ 0
2t² - 40t + 128 = 0
t² - 20t + 64 = 0
D = 400 - 256 = 144
t1 = (20 - 12)/2 = 4
t2 = (20 + 12)/2 = 16
x1 = -2
x2 = 2
x3 = -4
x4 = 4
0 votes
Thanks 1
MIA1k
128+2x^4-40x^2=0|:2
x^4-20x^2+64=0
x1x2=64
x1+x2=20
x1=16
x2=4
или
Д1=10*10-64=100-64=36=6^2
х1=10+6=16
x2=10-6=4
0 votes
Thanks 1
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Answers & Comments
Verified answer
X² = t ≥ 02t² - 40t + 128 = 0
t² - 20t + 64 = 0
D = 400 - 256 = 144
t1 = (20 - 12)/2 = 4
t2 = (20 + 12)/2 = 16
x1 = -2
x2 = 2
x3 = -4
x4 = 4
x^4-20x^2+64=0
x1x2=64
x1+x2=20
x1=16
x2=4
или
Д1=10*10-64=100-64=36=6^2
х1=10+6=16
x2=10-6=4