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Sagot :
[tex]|-3-5x|-8<-2\\
|-3-5x|<6\\
-3-5x<6 \wedge -3-5x>-6\\
-5x<9 \wedge -5x>-3\\
x>-\dfrac{9}{5} \wedge x<\dfrac{3}{5}\\
x\in\left(-\dfrac{9}{5},\dfrac{3}{5}\right)[/tex]
The hard way:
[tex]\left| -3-5x \right| -8<-2\\ \\ \left| -3-5x \right| <6\\ \\ { \left( -3-5x \right) }^{ 2 }<{ 6 }^{ 2 }\\ \\ \left( -3-5x \right) \left( -3-5x \right) <36[/tex]
[tex]\\ \\ 9+15x+15x+25{ x }^{ 2 }<36\\ \\ 25{ x }^{ 2 }+30x+9<36\\ \\ 25{ x }^{ 2 }+30x-27<0\\ \\ Say\quad f\left( x \right) =25{ x }^{ 2 }+30x-27,\\ \\ and\quad that\quad f\left( x \right) =0[/tex]
[tex]\\ \\ 25{ x }^{ 2 }+30x-27=0\\ \\ 25{ x }^{ 2 }+30x=27\\ \\ { x }^{ 2 }+\frac { 30 }{ 25 } x=\frac { 27 }{ 25 } \\ \\ { x }^{ 2 }+\frac { 6 }{ 5 } x=\frac { 27 }{ 25 } \\ \\ { \left( x+\frac { 3 }{ 5 } \right) }^{ 2 }-{ \left( \frac { 3 }{ 5 } \right) }^{ 2 }=\frac { 27 }{ 25 }[/tex]
[tex]\\ \\ { \left( x+\frac { 3 }{ 5 } \right) }^{ 2 }=\frac { 36 }{ 25 } \\ \\ x+\frac { 3 }{ 5 } =\pm \frac { 6 }{ 5 } \\ \\ x=-\frac { 3 }{ 5 } \pm \frac { 6 }{ 5 }[/tex]
[tex]Therefore:\\ \\ x=\frac { 3 }{ 5 } \quad and\quad x=-\frac { 9 }{ 5 } \quad when\quad f\left( x \right) =0\\ \\ Now:\\ \\ f\left( x \right) <0,\\ \\[/tex]
When:
-9/5<x<3/5
[tex]\left| -3-5x \right| -8<-2\\ \\ \left| -3-5x \right| <6\\ \\ { \left( -3-5x \right) }^{ 2 }<{ 6 }^{ 2 }\\ \\ \left( -3-5x \right) \left( -3-5x \right) <36[/tex]
[tex]\\ \\ 9+15x+15x+25{ x }^{ 2 }<36\\ \\ 25{ x }^{ 2 }+30x+9<36\\ \\ 25{ x }^{ 2 }+30x-27<0\\ \\ Say\quad f\left( x \right) =25{ x }^{ 2 }+30x-27,\\ \\ and\quad that\quad f\left( x \right) =0[/tex]
[tex]\\ \\ 25{ x }^{ 2 }+30x-27=0\\ \\ 25{ x }^{ 2 }+30x=27\\ \\ { x }^{ 2 }+\frac { 30 }{ 25 } x=\frac { 27 }{ 25 } \\ \\ { x }^{ 2 }+\frac { 6 }{ 5 } x=\frac { 27 }{ 25 } \\ \\ { \left( x+\frac { 3 }{ 5 } \right) }^{ 2 }-{ \left( \frac { 3 }{ 5 } \right) }^{ 2 }=\frac { 27 }{ 25 }[/tex]
[tex]\\ \\ { \left( x+\frac { 3 }{ 5 } \right) }^{ 2 }=\frac { 36 }{ 25 } \\ \\ x+\frac { 3 }{ 5 } =\pm \frac { 6 }{ 5 } \\ \\ x=-\frac { 3 }{ 5 } \pm \frac { 6 }{ 5 }[/tex]
[tex]Therefore:\\ \\ x=\frac { 3 }{ 5 } \quad and\quad x=-\frac { 9 }{ 5 } \quad when\quad f\left( x \right) =0\\ \\ Now:\\ \\ f\left( x \right) <0,\\ \\[/tex]
When:
-9/5<x<3/5
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