b) rozwiąż nierówność f(x) ˃1
a)
f(x)=ax²+bx+1
2=a+b+1
-1=4a+2b+1
a+b=1
4a+2b=-2
-2a-2b=-2
----------------+
2a=-4
a=-2
{a=-2
{b=3
f(x)=-2x²+3x+1
b)
-2x²+3x+1>1
-2x²+3x>0
x(-2x+3)>0
x(-2x+3)=0
x=0 lub x=1,5
x∈(0;1,5)
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a)
f(x)=ax²+bx+1
2=a+b+1
-1=4a+2b+1
a+b=1
4a+2b=-2
-2a-2b=-2
4a+2b=-2
----------------+
2a=-4
a=-2
{a=-2
{b=3
f(x)=-2x²+3x+1
b)
-2x²+3x+1>1
-2x²+3x>0
x(-2x+3)>0
x(-2x+3)=0
x=0 lub x=1,5
x∈(0;1,5)