А) х2 = 36
х = 6
б) х2 = 2,25
х = 1,5
в) 3х2 = 0
х2 = 0÷3
х2 = 0
х = 0
г) х2 = -1
не имеет смысла,
т. к. а (т.е. -1) < 0 - такого быть не может
-5,4 ; -5,04 ; 5,04))))))))
X1+x2=16⇒x1=16-x2
(x1)²-(x2)²=128
(16-x2)²-(x2)²=128
256-32x2+(x2)²-(x2)²=128
-32x2=128-256
-32x2=-128
x2=-128/-32=4
x1=16-4=8
c=x1*x2=4*8=32
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Log5(2x-3) > log5(x+5)
<span>2x-3 > x+5
</span>x > 8
ОДЗ: {x>1,5
{х> -5 x∈ (1,5 ;+∞)
Ответ: х∈(8 ;+∞)