0000:00:00:00:00Start21Number 21ExplanationThe solution set of the inequality ∣x−5∣2−3∣x−5∣+2<0|x - 5|^2 - 3|x - 5| + 2 < 0∣x−5∣2−3∣x−5∣+2<0 is....(3,4)∪[6,7)(3,4) \cup [6,7)(3,4)∪[6,7)(3,4)∪(6,7)(3,4) \cup (6,7)(3,4)∪(6,7)(1,2)∪(3,4](1,2) \cup (3,4](1,2)∪(3,4](−∞,1)∪[6,∞)(-\infty, 1) \cup [6,\infty)(−∞,1)∪[6,∞)(−∞,2)∪(3,7)(-\infty, 2) \cup (3,7)(−∞,2)∪(3,7)ExplanationLet ∣x−5∣=p|x - 5| = p∣x−5∣=p, we can solve the inequality to become ∣x−5∣2−3∣x−5∣+2<0|x - 5|^2 - 3|x - 5| + 2 < 0∣x−5∣2−3∣x−5∣+2<0p2−3p+2<0p^2 - 3p + 2 < 0p2−3p+2<0(p−1)(p−2)<0(p - 1)(p - 2) < 0(p−1)(p−2)<0p=1∪p=2p = 1 \cup p = 2p=1∪p=2 Return to the original form p=1→∣x−5∣=1→x=6∪x=4p = 1 \rightarrow |x - 5| = 1 \rightarrow x = 6 \cup x = 4p=1→∣x−5∣=1→x=6∪x=4p=2→∣x−5∣=2→x=7∪x=3p = 2 \rightarrow |x - 5| = 2 \rightarrow x = 7 \cup x = 3p=2→∣x−5∣=2→x=7∪x=3 The number line is as follows Number Line SolutionThe solution of (p−1)(p−2)<0(p - 1)(p - 2) < 0(p−1)(p−2)<0 for p=∣x−5∣p = |x - 5|p=∣x−5∣ yields 3<x<43 < x < 43<x<4 or 6<x<76 < x < 76<x<7.+++−-−+++−-−+++333333444444666666777777 Thus it has two solutions, namely HP={3<x<4}∪{6<x<7}HP = \{3 < x < 4\} \cup \{6 < x < 7\}HP={3<x<4}∪{6<x<7}. If shortened to (3,4)∪(6,7)(3,4) \cup (6,7)(3,4)∪(6,7).
21Number 21ExplanationThe solution set of the inequality ∣x−5∣2−3∣x−5∣+2<0|x - 5|^2 - 3|x - 5| + 2 < 0∣x−5∣2−3∣x−5∣+2<0 is....(3,4)∪[6,7)(3,4) \cup [6,7)(3,4)∪[6,7)(3,4)∪(6,7)(3,4) \cup (6,7)(3,4)∪(6,7)(1,2)∪(3,4](1,2) \cup (3,4](1,2)∪(3,4](−∞,1)∪[6,∞)(-\infty, 1) \cup [6,\infty)(−∞,1)∪[6,∞)(−∞,2)∪(3,7)(-\infty, 2) \cup (3,7)(−∞,2)∪(3,7)ExplanationLet ∣x−5∣=p|x - 5| = p∣x−5∣=p, we can solve the inequality to become ∣x−5∣2−3∣x−5∣+2<0|x - 5|^2 - 3|x - 5| + 2 < 0∣x−5∣2−3∣x−5∣+2<0p2−3p+2<0p^2 - 3p + 2 < 0p2−3p+2<0(p−1)(p−2)<0(p - 1)(p - 2) < 0(p−1)(p−2)<0p=1∪p=2p = 1 \cup p = 2p=1∪p=2 Return to the original form p=1→∣x−5∣=1→x=6∪x=4p = 1 \rightarrow |x - 5| = 1 \rightarrow x = 6 \cup x = 4p=1→∣x−5∣=1→x=6∪x=4p=2→∣x−5∣=2→x=7∪x=3p = 2 \rightarrow |x - 5| = 2 \rightarrow x = 7 \cup x = 3p=2→∣x−5∣=2→x=7∪x=3 The number line is as follows Number Line SolutionThe solution of (p−1)(p−2)<0(p - 1)(p - 2) < 0(p−1)(p−2)<0 for p=∣x−5∣p = |x - 5|p=∣x−5∣ yields 3<x<43 < x < 43<x<4 or 6<x<76 < x < 76<x<7.+++−-−+++−-−+++333333444444666666777777 Thus it has two solutions, namely HP={3<x<4}∪{6<x<7}HP = \{3 < x < 4\} \cup \{6 < x < 7\}HP={3<x<4}∪{6<x<7}. If shortened to (3,4)∪(6,7)(3,4) \cup (6,7)(3,4)∪(6,7).
