Рационални неравенства - задачи с решения

Решение:
[tex]\frac{5}{x+8}<0[/tex]

Неравенството е вярно само, когато [tex]x+8<0[/tex]
Така [tex]x<-8[/tex]

Решението е [tex]x\in \left( -\infty ,-8\right)[/tex]

Решение:
[tex]\frac{10}{2x+5}\geq 0[/tex]

Уравнението е възможно когато [tex]2x+5>0[/tex]

Тогава [tex]x>-\frac{5}{2}[/tex]

Решението е интервалът [tex]x\in \left( -\frac{5}{2},\infty \right)[/tex]

Решение:
Трябва да решим [tex]x^{2}-1<0[/tex]
Така [tex]\left( x+1\right) \left( x-1\right) <0[/tex]

Трябва да анализираме следната таблица от интервали

[tex]\begin{array}{cccccc} & \left( -\infty ,-1\right) & -1 & \left( -1,1\right) & 1 & \left( 1,+\infty \right) \\ \left( x+1\right) & \left( -\right) & 0 & \left( +\right) & 2 & \left( +\right) \\ \left( x-1\right) & \left( -\right) & -2 & \left( -\right) & 0 & \left( +\right) \\ \left( x+1\right) \left( x-1\right) & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}[/tex]

Решението на [tex]\frac{1}{x^{2}-1}<0[/tex] е [tex]x\in \left( -1,1\right)[/tex]

Решение:
[tex]\frac{x-3}{x+2}<0[/tex]

Корените са [tex]x=3\qquad x=-2[/tex]

Таблицата с интервалите изглежда така:

[tex]\begin{array}{cccccc} & \left( -\infty ,-2\right) & -2 & \left( -2,3\right) & 3 & \left( 3,+\infty \right) \\ \left( x+2\right) & \left( -\right) & 0 & \left( +\right) & 5 & \left( +\right) \\ \left( x-3\right) & \left( -\right) & -5 & \left( -\right) & 0 & \left( +\right) \\ \frac{x-3}{x+2} & \left( +\right) & \notin & \left( -\right) & 0 & \left( +\right) \end{array}[/tex]

Решението е [tex]x\in \left( -2,3\right)[/tex]

Решение:
[tex]\frac{x+5}{x}\geq 0[/tex]
Корените са [tex]x=-5\qquad x=0[/tex]
и сега да анализираме таблицата

[tex]\begin{array}{cccccc} & \left( -\infty ,-5\right) & -5 & \left( -5,0\right) & 0 & \left( 0,+\infty \right) \\ \left( x+5\right) & \left( -\right) & 0 & \left( +\right) & 5 & \left( +\right) \\ x & \left( -\right) & -5 & \left( -\right) & 0 & \left( +\right) \\ \frac{x+5}{x} & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}[/tex]

Решението е [tex]x\in (-\infty ,-5]\cup \left( 0,+\infty \right) [/tex]

Решение:
[tex]\frac{2x+6}{x-3}\leq 0[/tex]
Корените са [tex]x=-3\qquad x=3[/tex]

Нека да разгледаме таблицата

[tex]\begin{array}{cccccc} & \left( -\infty ,-3\right) & -3 & \left( -3,3\right) & 3 & \left( 3,+\infty \right) \\ 2x+6 & \left( -\right) & 0 & \left( +\right) & 12 & \left( +\right) \\ x-3 & \left( -\right) & -6 & \left( -\right) & 0 & \left( +\right) \\ \frac{2x+6}{x-3} & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}[/tex]

Решението е [tex]x\in \lbrack -3,3)[/tex]

Решение:
[tex]\frac{3x-1}{x+2}>0[/tex]
Корените са [tex]x=\frac{1}{3}\qquad x=-2[/tex]

да анализираме следните интервали

[tex]\begin{array}{cccccc} & \left( -\infty ,-2\right) & -2 & \left( -2,\frac{1}{3}\right) & \frac{1}{3} & \left( \frac{1}{3},+\infty \right) \\ 3x-1 & \left( -\right) & -7 & \left( -\right) & 0 & \left( +\right) \\ x+2 & \left( -\right) & 0 & \left( +\right) & \frac{7}{3} & \left( +\right) \\ \frac{3x-1}{x+2} & \left( +\right) & \notin & \left( -\right) & 0 & \left( +\right) \end{array}[/tex]

Решението е [tex]x\in \left( -\infty ,-2\right) \cup \left( \frac{1}{3},+\infty \right)[/tex]

Решение:
Трябва първо да преведем под общ знаменател.
[tex]\frac{x+1}{x-1}+2=\frac{x+1+2x-2}{x-1}>0\Longrightarrow \frac{3x-1}{x-1}>0[/tex]

Корените са [tex]x=\frac{1}{3}\qquad x=1[/tex]

И сега трябва да анализираме интервалите между корените

[tex]\begin{array}{cccccc} & \left( -\infty ,\frac{1}{3}\right) & \frac{1}{3} & \left( \frac{1}{3},1\right) & 1 & \left( 1,+\infty \right) \\ 3x-1 & \left( -\right) & 0 & \left( +\right) & 2 & \left( +\right) \\ x-1 & \left( -\right) & -\frac{2}{3} & \left( -\right) & 0 & \left( +\right) \\ \frac{3x-1}{x-1} & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}[/tex]

Решението е [tex]x\in \left( -\infty ,\frac{1}{3}\right) \cup \left(1,+\infty \right)[/tex]

Решение:
[tex]\frac{x-2}{x+3}\leq 1[/tex]

Първо прехвърляме всичко отляво, за да получим нула отдясно, и привеждаме под общ знаменател.
[tex]\frac{x-2}{x+3}-1\leq 0[/tex]

[tex]\frac{x-2-\left( x+3\right) }{x+3}\leq 0\Longrightarrow \frac{-5}{x+3}\leq 0[/tex]

Числителят е константата [tex]-5[/tex], затова знакът на дробта зависи само от знаменателя, а [tex]x=-3[/tex] се изключва.

[tex]\begin{array}{cccc} & \left( -\infty ,-3\right) & -3 & \left( -3,+\infty \right) \\ -5 & \left( -\right) & -5 & \left( -\right) \\ x+3 & \left( -\right) & 0 & \left( +\right) \\ \frac{-5}{x+3} & \left( +\right) & \notin & \left( -\right) \end{array}[/tex]

Дробта е отрицателна само когато [tex]x+3>0[/tex]

Решението е [tex]x\in \left( -3,+\infty \right)[/tex]

Решение:
Прехвърляме -2 отляво и привеждаме под общ знаменател.
[tex]\frac{2x-3}{5x+2}\geq -2[/tex]
[tex]\frac{2x-3}{5x+2}+2\geq 0[/tex]

Така [tex]\frac{2x-3+10x+4}{5x+2}\geq 0\Longrightarrow \frac{12x+1}{5x+2}\geq 0[/tex]
и намираме корените

[tex]x=-\frac{1}{12}\qquad x=-\frac{2}{5}[/tex]

Да анализираме таблицата

[tex]\begin{array}{cccccc} & \left( -\infty ,-\frac{2}{5}\right) & -\frac{2}{5} & \left( -\frac{2}{5},-\frac{1}{12}\right) & -\frac{1}{12} & \left( -\frac{1}{12},+\infty \right) \\ 12x+1 & \left( -\right) & -\frac{19}{5} & \left( -\right) & 0 & \left( +\right) \\ 5x+2 & \left( -\right) & 0 & \left( +\right) & \frac{19}{12} & \left( +\right) \\ \frac{12x+1}{5x+2} & \left( +\right) & \notin & \left( -\right) & 0 & \left( +\right) \end{array}[/tex]

Така решението е [tex]x\in \left( -\infty ,-\frac{2}{5}\right) \cup \lbrack -\frac{1}{12},+\infty )[/tex]

Решение:
[tex]\frac{3x-1}{2x-1}<-4[/tex]

[tex]\frac{3x-1}{2x-1}+4<0[/tex]

[tex]\frac{3x-1+8x-4}{2x-1}<0[/tex]

[tex]\frac{11x-5}{2x-1}<0[/tex]

[tex]x=\frac{5}{11}\qquad x=\frac{1}{2}[/tex]

[tex]\begin{array}{cccccc} & \left( -\infty ,\frac{5}{11}\right) & \frac{5}{11} & \left( \frac{5}{11},% \frac{1}{2}\right) & \frac{1}{2} & \left( \frac{1}{2},+\infty \right) \\ 11x-5 & \left( -\right) & 0 & \left( +\right) & \frac{1}{2} & \left( +\right) \\ 2x-1 & \left( -\right) & -\frac{1}{11} & \left( -\right) & 0 & \left( +\right) \\ \frac{11x-5}{2x-1} & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}[/tex]

[tex]x\in \left( \frac{5}{11},\frac{1}{2}\right) [/tex]

Решение:
Първо трябва да прехвърлим всичко отляво и да преобразуваме в един рационален израз.
[tex]x\leq 3-\frac{6}{x+2}\Longrightarrow x-3+\frac{6}{x+2}\leq 0[/tex]

[tex]\frac{\left( x-3\right) \left( x+2\right) +6}{x+2}\leq 0[/tex]
Така [tex]\frac{x^{2}-x}{x+2}\leq 0[/tex] [tex]\Longrightarrow \frac{x(x-1)}{x+2}\leq 0[/tex]

Корените са [tex]x=0, x=1[/tex] и [tex]x=-2[/tex]

[tex]\begin{array}{ccccccc} & \left( -\infty ,-2\right) & -2 & \left( -2,0\right) & 0 & \left( 0,1\right) & \left( 1,\infty \right) \\ x & \left( -\right) & -2 & \left( -\right) & 0 & \left( +\right) & \left( +\right) \\ \left( x-1\right) & \left( -\right) & -3 & \left( -\right) & -1 & \left( -\right) & \left( +\right) \\ \left( x+2\right) & \left( -\right) & 0 & (+) & 2 & \left( +\right) & \left( +\right) \\ \frac{x(x-1)}{x+2} & \left( -\right) & \notin & \left( +\right) & 0 & \left( -\right) & \left( +\right) \end{array}[/tex]

Решението е [tex]x\in \left( -\infty ,-2\right) \cup \left[ 0,1\right][/tex]

Решение:
Корените са [tex]x=0, x=1[/tex] и [tex]x=-5[/tex]

Нека да анализираме таблицата

[tex]\begin{array}{cccccccc} & \left( -\infty ,-5\right) & -5 & \left( -5,0\right) & 0 & \left( 0,1\right) & 1 & \left( 1,\infty \right) \\ x & \left( -\right) & -5 & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) \\ x-1 & \left( -\right) & -6 & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) \\ x+5 & \left( -\right) & 0 & \left( +\right) & 5 & \left( +\right) & 6 & \left( +\right) \\ \frac{x\left( x-1\right) }{x+5} & \left( -\right) & \notin & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}[/tex]

Така [tex]\frac{x\left( x-1\right) }{x+5}\geq 0[/tex], когато [tex]x\in (-5,0]\cup \lbrack1,\infty)[/tex]

Решение:
Корените са [tex]x=0, x=1[/tex] и [tex] x=-1[/tex] и анализираме

[tex]\begin{array}{cccccccc} & \left( -\infty ,-1\right) & -1 & \left( -1,0\right) & 0 & \left( 0,1\right) & 1 & \left( 1,\infty \right) \\ x & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) \\ 1+x & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) & 2 & \left( +\right) \\ 1-x & \left( +\right) & -2 & \left( +\right) & 1 & \left( +\right) & 0 & \left( -\right) \\ \frac{\left( 1+x\right) \left( 1-x\right) }{x} & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) & 0 & \left( -\right) \end{array}[/tex]

Уравнението [tex]\frac{\left( 1+x\right) \left( 1-x\right) }{x}<0[/tex] е изпълнено, когато [tex]x\in \left(-1,0\right) \cup \left( 1,\infty \right)[/tex]

Решение:
Първо трябва да прехвърлим всичко отляво и да преобразуваме в един рационален израз.
[tex]\frac{x^{2}-2x+3}{x+1}\leq 1\Longrightarrow \frac{x^{2}-2x+3}{x+1}-1\leq 0[/tex]

Така [tex]\frac{x^{2}-2x+3-x-1}{x+1}\leq 0\Longrightarrow \frac{x^{2}-3x+2}{x+1}\leq 0\Longrightarrow \frac{\left( x-1\right) \left( x-2\right) }{x+1}\leq 0[/tex]

Корените са [tex]x=-1, x=1[/tex] и [tex]x=2[/tex]

[tex]\begin{array}{cccccccc} & \left( -\infty ,-1\right) & -1 & \left( -1,1\right) & 1 & \left( 1,2\right) & 2 & \left( 2,\infty \right) \\ x-1 & \left( -\right) & -2 & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) \\ x-2 & \left( -\right) & -3 & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) \\ x+1 & \left( -\right) & 0 & \left( +\right) & 2 & \left( +\right) & 3 & \left( +\right) \\ \frac{\left( x-1\right) \left( x-2\right) }{x+1} & \left( -\right) & \notin & \left( +\right) & 0 & \left( -\right) & 0 & \left( +\right) \end{array}[/tex]

Неравенството [tex]\frac{\left( x-1\right) \left( x-2\right) }{x+1}\leq 0[/tex] е изпълнено, когато [tex]x\in \left( -\infty ,-1\right) \cup \left[ 1,2\right][/tex]

Решение:
[tex]\frac{x}{x^{2}-1}\geq 0[/tex]

Тогава [tex]\frac{x}{\left( x+1\right) \left(x-1\right) }\geq 0[/tex]

Корените са [tex]x=0, x=-1[/tex] и [tex]x=1[/tex]

Трябва да анализираме следните интервали:

[tex]\begin{array}{cccccccc} & \left( -\infty ,-1\right) & -1 & \left( -1,0\right) & 0 & \left( 0,1\right) & 1 & \left( 1,\infty \right) \\ x-1 & \left( -\right) & -2 & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) \\ x & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) \\ x+1 & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) & 2 & \left( +\right) \\ \frac{x}{\left( x+1\right) \left( x-1\right) } & \left( -\right) & \notin & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}[/tex]

Така [tex]\frac{x}{\left( x+1\right) \left( x-1\right) }\geq 0[/tex], когато [tex]x\in (-1,0]\cup (1,\infty )[/tex]

Решение:
[tex]\frac{\left( x^{2}-1\right) }{\left( x^{2}-4\right) }\geq 0[/tex]
Трябва да намерим корените на

[tex]\left( x^{2}-1\right) =\left( x-1\right) \left(x+1\right)[/tex]
[tex]\left( x^{2}-4\right) =\left( x-2\right) \left( x+2\right) \Longrightarrow x=-2,-1,1,2[/tex] са корените.

Анализираме интервалите

[tex]\begin{array}{cccccccccc} & \left( -\infty ,-2\right) & -2 & \left( -2,-1\right) & -1 & \left( -1,1\right) & 1 & \left( 1,2\right) & 2 & \left( 2,\infty \right) \\ x-2 & \left( -\right) & -4 & \left( -\right) & -3 & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) \\ x-1 & \left( -\right) & -3 & \left( -\right) & -2 & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) \\ x+1 & \left( -\right) & -1 & \left( -\right) & 0 & \left( +\right) & 2 & \left( +\right) & 3 & \left( +\right) \\ x+2 & \left( -\right) & 0 & \left( +\right) & 1 & \left( +\right) & 3 & \left( +\right) & 4 & \left( +\right) \\ \frac{\left( x^{2}-1\right) }{\left( x^{2}-4\right) } & \left( +\right) & \notin & \left( -\right) & 0 & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}[/tex]

Тогава [tex]\frac{\left( x^{2}-1\right) }{\left( x^{2}-4\right) }\geq 0[/tex], когато [tex]x\in \left( -\infty ,-2\right) \cup \left[ -1,1\right] \cup \left( 2,\infty\right)[/tex]

Решение:
[tex]\frac{2}{x-2}+\frac{3}{x-3}\leq 0[/tex]

Трябва да преобразуваме лявата страна в един рационален израз.

[tex]\frac{2}{x-2}+\frac{3}{x-3}=\frac{2\left( x-3\right) +3\left( x-2\right) }{\left( x-2\right) \left( x-3\right) }=\frac{5x-12}{\left( x-2\right) \left( x-3\right) }\leq 0[/tex]

Корените са [tex]x=\frac{12}{5}, x=2[/tex] и [tex]x=3[/tex] и сега трябва да намерим решението, анализирайки таблицата:

[tex]\begin{array}{cccccccc} & \left( -\infty ,2\right) & 2 & \left( 2,\frac{12}{5}\right) & \frac{12}{5% } & \left( \frac{12}{5},3\right) & 3 & \left( 3,\infty \right) \\ 5x-12 & \left( -\right) & -2 & \left( -\right) & 0 & \left( +\right) & 3 & \left( +\right) \\ x-2 & \left( -\right) & 0 & \left( +\right) & \frac{2}{5} & \left( +\right) & 1 & \left( +\right) \\ x-3 & \left( -\right) & -1 & \left( -\right) & -\frac{3}{5} & \left( -\right) & 0 & \left( +\right) \\ \frac{5x-12}{\left( x-2\right) \left( x-3\right) } & \left( -\right) & \notin & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}[/tex]

Решението е [tex]x\in \left( -\infty ,2\right) \cup \lbrack \frac{12}{5},3)[/tex]

Решение:
Когато разложим на множители [tex]x^{2}+2x-3=\left( x+3\right) \left( x-1\right)[/tex] и [tex]x^{2}-1=\left( x-1\right) \left( x+1\right)[/tex]

получаваме 3 корена [tex]x=-3, x=-1[/tex] и [tex]x=1[/tex]
така

[tex]\frac{x^{2}+2x-3}{x^{2}-1}=\frac{\left( x+3\right) \left( x-1\right) }{\left( x-1\right) \left( x+1\right)}=\frac{\left( x+3\right) }{\left(x+1\right) }<0[/tex]
тогава

[tex]\begin{array}{cccccc} & \left( -\infty ,-3\right) & -3 & \left( -3,-1\right) & -1 & \left( -1,+\infty \right) \\ \left( x+3\right) & \left( -\right) & 0 & \left( +\right) & 2 & \left( +\right) \\ \left( x+1\right) & \left( -\right) & -2 & \left( -\right) & 0 & \left( +\right) \\ \frac{\left( x+3\right) }{\left( x+1\right) } & \left( +\right) & 0 & \left( -\right) & \notin & \left( +\right) \end{array}[/tex]

Решението е [tex]x\in \left( -3,-1\right) [/tex]