а) ОДЗ: [tex]x^2+8x-9\neq 0[/tex]
[tex]D=b^2-4ac = 64 - 4*(-9)=64+36=100\\x_{1}=\frac{-b-\sqrt{D} }{2a}=\frac{-8-10}{2}=-9\\x_{2}= \frac{-b+\sqrt{D} }{2a}=\frac{-8+10}{2}=1[/tex]
х∈(-∞; -9)U(-9; 1)U(1; +∞)
б) ОДЗ: 5x + 1 > 0
5x > -1
x > -1/5
х∈(-1/5; +∞)
в) ОДЗ: [tex]\left \{ {{9-3x\geq 0} \atop {2x+5\geq 0}} \right.[/tex]
[tex]\left \{ {{x\leq 3} \atop {x\geq 2,5}} \right.[/tex]
х∈[2,5; 3]
г) ОДЗ:
[tex]\left \{ {{-x+2\geq 0} \atop {6-2x > 0}} \right. \\[/tex]
[tex]\left \{ {{-x+2\geq 0} \atop {6-2x > 0}} \right. \\\left \{ {{x\leq 2} \atop {x\leq 3}} \right.[/tex]
х∈(-∞; 2]
д) ОДЗ: [tex]x\neq 0[/tex]
2x+4>0
2x>-4
x>-2
х∈(-2; 0)U(0; +∞)
е) ОДЗ:
[tex]\left \{ {{4x+2\geq 0} \atop {x+2\neq 0}} \right.[/tex]
[tex]\left \{ {{x\geq -\frac{1}{2} }\\ \atop {x\neq -2}} \right.[/tex]
х∈(1/2; +∞)
Copyright © 2025 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
а) ОДЗ: [tex]x^2+8x-9\neq 0[/tex]
[tex]D=b^2-4ac = 64 - 4*(-9)=64+36=100\\x_{1}=\frac{-b-\sqrt{D} }{2a}=\frac{-8-10}{2}=-9\\x_{2}= \frac{-b+\sqrt{D} }{2a}=\frac{-8+10}{2}=1[/tex]
х∈(-∞; -9)U(-9; 1)U(1; +∞)
б) ОДЗ: 5x + 1 > 0
5x > -1
x > -1/5
х∈(-1/5; +∞)
в) ОДЗ: [tex]\left \{ {{9-3x\geq 0} \atop {2x+5\geq 0}} \right.[/tex]
[tex]\left \{ {{x\leq 3} \atop {x\geq 2,5}} \right.[/tex]
х∈[2,5; 3]
г) ОДЗ:
[tex]\left \{ {{-x+2\geq 0} \atop {6-2x > 0}} \right. \\[/tex]
[tex]\left \{ {{-x+2\geq 0} \atop {6-2x > 0}} \right. \\\left \{ {{x\leq 2} \atop {x\leq 3}} \right.[/tex]
х∈(-∞; 2]
д) ОДЗ: [tex]x\neq 0[/tex]
2x+4>0
2x>-4
x>-2
х∈(-2; 0)U(0; +∞)
е) ОДЗ:
[tex]\left \{ {{4x+2\geq 0} \atop {x+2\neq 0}} \right.[/tex]
[tex]\left \{ {{x\geq -\frac{1}{2} }\\ \atop {x\neq -2}} \right.[/tex]
х∈(1/2; +∞)