10.3: 理性不平等
- Page ID
- 171341
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)求解有理不等式需要找到分子和分母的零，然后使用这些值调查数字线上的解集区域。
求解不等式并用区间表示法写出解集：
- \(\dfrac{x − 1}{x + 1} ≥ 0\)
- \(\dfrac{2x − 3}{x + 1} ≤ 0\)
- \(\dfrac{x + 2}{x − 2} ≥ 0\)
解决方案
- \(\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 &\ text {找到分子的零}\\ &x + 1 = 0, x =\; −1 &\ text {找到分母的零}\ end {array}\)
\(\begin{array} &&\text{For } x < −1, \text{ choose } x = −2. \;\;\dfrac{−2 − 1}{−2 + 1} = \dfrac{−3}{−1} = 3 ≥ 0 \\ &\text{Replacing \(-2\)表示答案中的\(x\)结果\(3\)，该结果大于或等于\(0\)。 \(x < −1\)该区域包含在解决方案集中。}\\ [0.25in] &\ text {For} −1 < x < 1，\ text {choose} x = 0。\;\ dfrac {0 − 1} {0 + 1} = −1 < 0\\ &\ text {替换\(0\)答案中的\(x\)结果\(-1\)，小于\(0\)，不满足问题中给定的不等式。}\\ &\ text {此区域\(−1 < x < 1\)不包括在解集中。}\\ [0.25in] &\ text {For} x > 1，\ text {choose} x = 2.\;\;\ dfrac {2 + 1} {2 + 1} =\ dfrac {1} ≥ 0\\ &\ text {替换\(2\)答案中的\(x\)结果\(\dfrac{1}{3}\)，哪个是大于或等于\(0\)。 \(x > 1\)该区域包含在解集中。}\\ [0.25in] & (−∞, −1) (1, ∞)\\ &\ text {用区间表示法书写的最终答案（有关更多详细信息，请参阅区间表示法部分）。} \ end {array}\)
- \(\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 &\ text {找到分子的零}\\ &x + 1 = 0,\; x = −1 &\ text {找到分母的零}\ end {array}\)
\(\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 \(-2\)表示答案中的\(x\)结果大于\(\dfrac{7}{3}\)\(0\)，未满足问题中给定的不等式。}\\ &\ text {此区域\(x < −1\)不在解集合中。}\\ [0.25in] &\ text {For} −1 < x < 1.5，\ text {choose} x = 0。\;\ dfrac {2 (0) −3} {0 − 1} =\ dfrac {−3} {−1} = 3 ≥ 0\\ &\ text {替换\(0\)答案中的\(x\)结果\(3\)，大于或等于\(0\)，这不是问题所要求的。}\\ &\ text {解集中不包括该区域\(−1 < x < 1.5\)。}\\ [0.25in] &\ text {For} x > 1，\ text {choose} x = 2。\;\ dfrac {2 (2) − 3} {2 − 1} =\ dfrac {1} {1} = 1 ≥ 0\\ &\ text {替换\(2\)答案中的\(x\)结果\(1\)，大于或等于\(0\)。 该区域\(x > 1\)不包含在解决方案集中。}\\ [0.25in] &\\ &\ text {这个问题没有解决方案。 \(\dfrac{2x − 3}{x + 1}\)永远不会小于或等于\(0\)。} \ end {array}\)
- \(\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 = −2 &\ text {找到分子的零}\\ &x − 2 = 0,\;\; x = 2 &\ text {找到分母的零}\ end {array}\)
\(\begin{array} &&\text{For } x < −2, \text{ choose } x = −3. \dfrac{−3 + 2}{−3 − 2} = \dfrac{−1}{−5} = \dfrac{1}{5} ≥ 0 \\ &\text{Replacing \(-3\)表示答案中的\(x\)结果\(\dfrac{1}{5}\)，该结果大于或等于\(0\)。 \(x < −2\)该区域包含在解决方案集中。}\\ [0.25in] &\ text {For} −2 < x < 2，\ text {choose} x = 0。\;\ dfrac {0 + 2} {0 − 2} = −1 < 0\\ &\ text {替换\(0\)答案中的\(x\)结果\(-1\)，小于\(0\)，不满足问题中给定的不等式。}\\ &\ text {此区域\(−2 < x < 2\)不包含在解集中。}\\ [0.25in] &\ text {For} x > 2，\ text {choose} x = 3.\;\ dfrac {3 + 2} {3 − 2} =\ dfrac {5} {1} = 5 ≥ 0\\ &\ text {替换\(3\)\(x\)结果为回答\(5\)，它大于或等于\(0\)。 \(x > 2\)该区域包含在解集中。}\\ [0.25in] & (−∞, −2) (2, ∞)\\ &\ text {用区间表示法书写的最终答案（更多详情请参阅区间表示法部分）。} \ end {array}\)
- \(\dfrac{x + 3}{x − 2} ≥ 0\)
- \(\dfrac{x − 2}{x − 1} ≤ 0\)
- \(\dfrac{2x − 1}{x + 2} ≤ 0\)
- \(\dfrac{2x − 3}{x + 1} ≥ 0\)