Skip to main content
Global

10.3: Ukosefu wa usawa wa busara

  1. Last updated
  2. Save as PDF
  • Page ID
    164817

    • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Kutatua usawa wa busara unahusisha kutafuta zero za nambari na denominator, kisha kutumia maadili haya kuchunguza mikoa ya kuweka ufumbuzi kwenye mstari wa namba.

    Mfano Template:index

    Tatua usawa na uandike seti za ufumbuzi katika notation ya muda:

    1. \(\dfrac{x − 1}{x + 1} ≥ 0\)
    2. \(\dfrac{2x − 3}{x + 1} ≤ 0\)
    3. \(\dfrac{x + 2}{x − 2} ≥ 0\)
    Suluhisho
    1. \(\begin{array} &&\dfrac{x − 1}{x + 1} ≥ 0 &\text{Example problem} \\ &\dfrac{x − 1}{x + 1} ≥ 0 &\text{The quotient must be greater than or equal to \(0\).}\\ &x- 1 = 0,\; x = 1 &\ maandishi {Pata zero za namba}\\ &x + 1 = 0, x =\; -1 &\ maandishi {Pata zero za denominator}\ mwisho {safu}\)
    clipboard_ec11760d50b919ddd83bc64d27535e4b7.png
    Zero hugawanya mstari wa namba katika\(3\) mikoa\(x < −1\),\(−1 < x < 1\),\(x > 1\)

    \(\begin{array} &&\text{For } x < −1, \text{ choose } x = −2. \;\;\dfrac{−2 − 1}{−2 + 1} = \dfrac{−3}{−1} = 3 ≥ 0 \\ &\text{Replacing \(-2\)kwa\(x\) matokeo katika jibu\(3\), ambayo ni kubwa kuliko au sawa na\(0\). Eneo hili\(x < −1\) linajumuishwa katika seti ya ufumbuzi.}\\ [0.25in] &\ maandishi {Kwa} -1 <x <1,\ maandishi {kuchagua} x = 0.\;\;\ dfrac {0} {0 + 1} =\ dfrac {-1} {1} {1} = -1 <0\\ &\ maandishi {Kubadilisha\(0\)\(x\) matokeo katika jibu\(-1\), ambayo ni chini ya\(0\), si kutimiza kutofautiana kutokana na tatizo.}\\ &\ maandishi {Eneo hili\(−1 < x < 1\) linatengwa na kuweka suluhisho.}\\ [0.25in] &\ maandishi {Kwa} x > 1,\ maandishi {kuchagua} x = 2.\;\;\ dfrac {2 ∙ 1} {2 + 1} =\ dfrac {1} {1} {3} ≥ 0\\ &\ maandishi {Kubadilisha\(2\)\(x\) matokeo katika jibu\(\dfrac{1}{3}\), ambayo ni kubwa kuliko au sawa na\(0\). Mkoa huu\(x > 1\) umejumuishwa katika seti ya ufumbuzi.}\\ [0.25in] & (-Δ, -1) (1, Δ)\\ &\ maandishi {Jibu la mwisho lililoandikwa katika nukuu za muda (angalia sehemu ya Nukuu ya Muda kwa maelezo zaidi).} \ mwisho {safu}\)

    1. \(\begin{array} &&\dfrac{2x − 3}{x + 1} ≤ 0 &\text{Example problem} \\ &\dfrac{2x − 3}{x + 1} ≤ 0 &\text{The quotient must be less than or equal to \(0\).}\\ &2x - 3 = 0,\; x = 1.5 &\ maandishi {Pata zero za namba}\\ &x + 1 = 0,\; x = -1 &\ maandishi {Pata zero za denominator}\ mwisho {safu}\)
    clipboard_eb1a3fd2706ba7b03d6acdeb3ad51ec07.png
    Zero hugawanya mstari wa namba katika\(3\) mikoa\(x < −1\),\(−1 < x < 1.5\),\(x > 1.5\)

    \(\begin{array} &&\text{For } x < −1, \text{ choose } x = −2. \;\; \dfrac{2(−2) − 3}{−2 − 1} = \dfrac{−7}{−3} = \dfrac{7}{3} ≥ 0 \\ &\text{Replacing \(-2\)kwa\(x\) matokeo katika jibu\(\dfrac{7}{3}\), ambayo ni kubwa kuliko\(0\), si kutimiza usawa uliotolewa katika tatizo.}\\ &\ maandishi {Eneo hili\(x < −1\) limetengwa na kuweka suluhisho.}\\ [0.25in] &\ maandishi {Kwa} —1 <x <1.5,\ maandishi {chagua} x = 0.\;\\;\ dfrac {2 (0) ∙ 3} {0 - 1} =\ dfrac {-3} {-1} = 3 ≥ 0\\ &\ maandishi {Kubadilisha\(0\)\(x\) matokeo katika jibu\(3\), ambayo ni kubwa kuliko au sawa\(0\), ambayo si kile tatizo linaomba.}\\ &\ maandishi {Eneo hili\(−1 < x < 1.5\) linatengwa katika seti ya suluhisho.}\\ [0.25in] &\ maandishi {Kwa } x > 1,\ maandishi {kuchagua} x = 2.\;\;\ dfrac {2 (2) - 3} {2 ∙ 1} =\ dfrac {1} {1} = 1 ≥ 0\\ &\ maandishi {Kubadilisha\(2\)\(x\) matokeo katika jibu\(1\), ambayo ni kubwa kuliko au sawa\(0\). Eneo hili\(x > 1\) limetengwa katika seti ya suluhisho.}\\ [0.25in] &\\ &\ maandishi {Tatizo hili halina suluhisho. \(\dfrac{2x − 3}{x + 1}\)kamwe kuwa chini ya au sawa na\(0\).} \ mwisho {safu}\)

    1. \(\begin{array} &&\dfrac{x + 2}{x − 2} ≥ 0 &\text{Example problem} \\ &\dfrac{x + 2}{x − 2} ≥ 0 &\text{The quotient must be greater than or equal to \(0\).}\\ &x + 2 = 0,\;\; x = -1 &\ maandishi {Pata zero za namba}\\ &x- 2 = 0,\;\; x = 2 &\ maandishi {Pata zero za denominator}\ mwisho {safu}\)
    clipboard_e3374fd2eab55feb8c3dfe2d5ebd8d557.png
    Zero hugawanya mstari wa namba katika\(3\) mikoa\(x < −2\),\(−2 < x < 2\),\(x > 2\)

    \(\begin{array} &&\text{For } x < −2, \text{ choose } x = −3. \dfrac{−3 + 2}{−3 − 2} = \dfrac{−1}{−5} = \dfrac{1}{5} ≥ 0 \\ &\text{Replacing \(-3\)kwa\(x\) matokeo katika jibu\(\dfrac{1}{5}\), ambayo ni kubwa kuliko au sawa na\(0\). Mkoa huu\(x < −2\) umejumuishwa katika seti ya ufumbuzi.}\\ [0.25in] &\ maandishi {Kwa} -2 <x <2,\ maandishi {kuchagua} x = 0.\;\;\ dfrac {0 + 2} {0 δ 2} =\ dfrac {2} {¯ 2} = -1 <0\\ &\ maandishi {Kubadilisha\(0\)\(x\) matokeo katika jibu\(-1\), ambayo ni chini ya\(0\), si kutimiza kutofautiana kutokana na tatizo.}\\ &\ maandishi {Eneo\(−2 < x < 2\) hili halijumuishwa katika kuweka suluhisho.}\\ [0.25in] &\ maandishi {Kwa} x > 2,\ maandishi {kuchagua} x = 3.\;\;\ dfrac {3 + 2} {3 - 2} =\ dfrac {5} {1} = 5 ≥ 0\\ &\ maandishi {Kubadilisha\(3\) kwa\(x\) matokeo jibu\(5\), ambayo ni kubwa kuliko au sawa na\(0\). Mkoa huu\(x > 2\) umejumuishwa katika seti ya ufumbuzi.}\\ [0.25in] & (-Δ, -1) (2, Δ)\\ &\ maandishi {Jibu la mwisho lililoandikwa katika nukuu za muda (angalia sehemu ya Nukuu ya Muda kwa maelezo zaidi).} \ mwisho {safu}\)

    Zoezi Template:index
    1. \(\dfrac{x + 3}{x − 2} ≥ 0\)
    2. \(\dfrac{x − 2}{x − 1} ≤ 0\)
    3. \(\dfrac{2x − 1}{x + 2} ≤ 0\)
    4. \(\dfrac{2x − 3}{x + 1} ≥ 0\)