При 1+2y≥0;
2y≥-1
y≥-0,5
y∈[-0,5;+∞)
1) =0,04x^2-4xy+100y^2
3) =a^4-2a^2+1
Ответ:
Объяснение:
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3*log₁/₃(x)<log₁/₃9+log₁/₃3 ОДЗ: x>0
3*log₁/₃(x)<log₁/₃27
3*log₁/₃(x)<-log₃3³
3*log₁/₃(x)<-3 I·3
log₁/₃(x)<-1
x>(1/3)⁻¹
x>3.
5-11х= -2x²
2x²-11x+5=0
D=121-40=81
x<span>1=11-9/10=2/10=0.2
x2=11+9/10=20/10=2
</span>9x²<span>-24х= -16
</span>9x²-24x+16=0
k=12
D=144-144=0;D=0
x=24:18=4/3
6x²<span>-4=0
</span>6x²=4
x²=2/3
x=+-√2/3