1.
ОДЗ:
х+2≠0
x≠ -2
D(y)=(-∞; -2)U(-2; +∞)
2.
ОДЗ:
x²-5x-6≠0
x²-5x-6=0
D=(-5)²-4*(-6)=25+24=49
x₁=(5-7)/2= -1
x₂=(5+7)/2=8
x≠ -1 x≠8
D(y)=(-∞; -1)U(-1; 8)U(8; +∞)
3.
a) 8-x≥0
-x≥ -8
x≤8
b) x+2≥0
x≥ -2
D(y)=[-2; 8]
4.
ОДЗ:
a) 3+x≥0
x≥ -3
b) x² -25≠0
x≠ -5 x≠5
D(y)=[-3; 5)U(5; +∞)
5.
ОДЗ:
а) -3х+5≠0
-3х≠ -5
х≠ -5/3
х≠ -1 ²/₃
b) (-6x+x²-27)/(-3x+5) ≥0
Разложим на множители:
x² -6x-27=0
D=(-6)² -4*(-27)=36+108=144
x₁=(6-12)/2= -3
x₂=(6+12)/2=9
x²-6x-27=(x+3)(x-9)
Метод интервалов:
(x+3)(x-9)(-3x+5)≥0
-3(x+3)(x-9)(x-1 ²/₃)≥0
(x+3)(x-9)(x-1 ²/₃)≤0
x= -3 x=9 x=1 ²/₃
- + - +
------------ -3 ----------- 1 ²/₃ ----------- 9 ------------
\\\\\\\\\\\\\ \\\\\\\\\\\\\\
x= -4 - - - | -
x= 0 + - - | +
x= 2 + - + | -
x= 10 + + + | +
x∈(-∞; -3]U(1 ²/₃; 9]
D(y)=(-∞; -3]U(1 ²/₃; 9]
6.
ОДЗ:
а) -6x+x²-27≥0
x² -6x-27≥0
(x+3)(x-9)≥0
x= -3 x=9
+ - +
-------- -3 ----------- 9 ----------
\\\\\\\\\\ \\\\\\\\\\\\\
x∈(-∞; -3]U[9; +∞)
b) -3x+5>0
-3x> -5
x< 1 ²/₃
В итоге ОДЗ: x∈(-∞; -3]
D(y)=(-∞; -3]
7.
ОДЗ:
a) x-4≠0
x≠4
b) [(4-x)²(x+2)³(x-7)⁴]/(x-4) ≥0
[(x-4)²(x+2)³(x-7)⁴]/(x-4)≥0
(x-4)(x+2)³(x-7)⁴≥0
x=4 x= -2 x=7
+ - + +
-------- -2 ---------- 4------------ 7 -----------
\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x= -3 - - +| +
x=0 - + +| -
x=5 + + +| +
x=8 + + +| +
x∈(-∞; -2]U[4; +∞)
В итоге х∈(-∞; -2]U(4; +∞)
D(y)=(-∞; -2]U(4; +∞)