e)
f(x)=- (5x-7)/√(4-(x+1)²)
4-(x+1)² > 0
[2-(x+1)][2+(x+1)] > 0
(2-x-1)(2+x+1) > 0
(1-x)(3+x) > 0
-(x-1)(x+3) > 0 |:(-1)
(x-1)(x+3) < 0
x-1=0 ∨ x+3=0
x=1 ∨ x=-3
x∈(-3,1)
f)
f(x)=(2x²-3)/√(25x²-40x+16)
25x²-40x+16 > 0
Δ=(-40)²-4·25·16=1600+1600=3200 , √Δ=√3200=40√2=40√2
x1=(40-40√2)/50
x1=(4-4√2)/5
x2=(40+40√2)/50
x2=(4+4√2)/5
x∈(-∞,(4-4√2)/5 ) ∪ ( (4+4√2)/5,∞)
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e)
f(x)=- (5x-7)/√(4-(x+1)²)
4-(x+1)² > 0
[2-(x+1)][2+(x+1)] > 0
(2-x-1)(2+x+1) > 0
(1-x)(3+x) > 0
-(x-1)(x+3) > 0 |:(-1)
(x-1)(x+3) < 0
x-1=0 ∨ x+3=0
x=1 ∨ x=-3
x∈(-3,1)
f)
f(x)=(2x²-3)/√(25x²-40x+16)
25x²-40x+16 > 0
Δ=(-40)²-4·25·16=1600+1600=3200 , √Δ=√3200=40√2=40√2
x1=(40-40√2)/50
x1=(4-4√2)/5
x2=(40+40√2)/50
x2=(4+4√2)/5
x∈(-∞,(4-4√2)/5 ) ∪ ( (4+4√2)/5,∞)