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zad 1
a) √(4x²-6)>0
4x²-6>0
Δ=-4*4*(-6)=-16*(-6)=96
√Δ=4√6
x₁=-4√6/8=-√6/2
x₂=√6/2
Df: x∈ (-∞;-√6/2)u(√6/2;+∞)
b) x²-6x≠0
x(x-6)≠0
x=0
x=6
Df: xeR\{0,6}
c) -x²+2x³>0
x²(-1+2x)>0
-1+2x=0
2x=1
x=1/2
Df: x∈ (1/2,+∞)
d) 3x²-x-4≠0
Δ=1-4*3*(-4)=1+48=49
√Δ=7
x₁=(1-7)/6=-6/6=-1
x₂=(1+7)/6=8/6=4/3
df: x∈R\{-1; 4/3}
zad 2
a)
1/2x-6>0 //*2
x-12>0
x>12
x∈(12,+∞)
b)
1/2x²-6≥0
Δ=-4*1/2*(-6)=-2*(-6)=12
√Δ=2√3
x₁=-2√3
x₂=2√3
Df: x∈(-∞;-2√3>u<2√3;+∞)
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zad 1
a) √(4x²-6)>0
4x²-6>0
Δ=-4*4*(-6)=-16*(-6)=96
√Δ=4√6
x₁=-4√6/8=-√6/2
x₂=√6/2
Df: x∈ (-∞;-√6/2)u(√6/2;+∞)
b) x²-6x≠0
x(x-6)≠0
x=0
x=6
Df: xeR\{0,6}
c) -x²+2x³>0
x²(-1+2x)>0
x=0
-1+2x=0
2x=1
x=1/2
Df: x∈ (1/2,+∞)
d) 3x²-x-4≠0
Δ=1-4*3*(-4)=1+48=49
√Δ=7
x₁=(1-7)/6=-6/6=-1
x₂=(1+7)/6=8/6=4/3
df: x∈R\{-1; 4/3}
zad 2
a)
1/2x-6>0 //*2
x-12>0
x>12
x∈(12,+∞)
b)
1/2x²-6≥0
Δ=-4*1/2*(-6)=-2*(-6)=12
√Δ=2√3
x₁=-2√3
x₂=2√3
Df: x∈(-∞;-2√3>u<2√3;+∞)
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