What's the difference between the solution of inequality and the existence of inequality? For example, is x greater than 1-B / a the same as x greater than 1-B / a? Good and quick answer,
Constant holds no matter what value x takes, for example, x ^ 2 > = 0
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- 1. In △ ABC, point D is the intersection of bisector of ∠ C and base ab. the following inequality is proved: CD & # 178; = AC · BC-AD · BD
- 2. Three inequalities are known: ① AB > 0. ② BC ad > 0. ③ C / a > D / b. taking two of them as conditions and the remaining one as conclusion, the number of correct propositions that can be formed is
- 3. If a > b, C
- 4. If A-B = 1 + √ 3, B-C = 1 - √ 3, find the value of 1 / (A & # 178; + B & # 178; + C & # 178; - AB AC BC)
- 5. If point C is the golden section of line AB, and AC
- 6. The position of the known real number AB on the number axis is shown in the figure: try to simplify: root (a-b) 178; - | a + B|
- 7. The positions of real numbers a and B on the number axis are shown in the figure. Try to simplify | a + B | + | A-B | + | ab| -----b--------0--------a--------》
- 8. The positions of real numbers a and B on the number axis are shown in the figure ────┴──┴─────┴─────┴─┴──> b -1 0 a 1 A.a+b>0 B.ab>0 C.a-b>0 D.|a|>|b|
- 9. If two points a and B on the number axis represent real numbers a and B respectively, the length of line AB is () A. a-bB. a+bC. |a-b|D. |a+b|
- 10. An inequality is a formula that uses (), (), or () to represent the size relationship
- 11. On the existence of inequalities For any real number x, the inequality │ x + 1 │ + X-2 │ > a holds, and the value range of real number a is obtained Analysis 2: using the absolute value inequality │ a │ - B │ < a ± B │ < a │ + B │ to solve the minimum value of F (x) = │ x + 1 │ + X-2 │ Let f (x) = x + 1 + X-2, x + 1 + X-2 = 3, f (x) min = 3. A < 3 Why can we do this?
- 12. Given C > 0, let P: function y = C ^ X be a decreasing function on R; Q: inequality x + | x-2c | >
- 13. It is proved that for any a, B, C, D ∈ R, there is an inequality (AC + BD) ≤ (a + b) (c + D),
- 14. It is proved that for any ABCD, there is always (AC + BD) & sup2; ≤ (A & sup2; + B & sup2;) (C & sup2; + D & sup2;)
- 15. Prove the inequality: (A & sup2; + B & sup2;) (C & sup2; + D & sup2;) ≥ (AC + BD) & sup2;
- 16. It is proved that for any a, B, C and D, they belong to R, and there is an inequality (AC + BD) ^ 2
- 17. The proof of inequality (AC + BD) & sup2; ≤ (A & sup2; + B & sup2;) (C & sup2; + D & sup2;)
- 18. There are four inequalities: (1) A2 + B2 + C2 ≥ AB + BC + AC; (2) a (1-A) ≤ 14; (3) BA + ab ≥ 2; (4) (A2 + B2) (C2 + D2) ≥ (AC + BD) 2; where () A. 1 B. 2 C. 3 d. 4
- 19. Three inequalities are known: (1) AB > 0, (2) C / a > D / B, (3) BC > ad If two of them are taken as the conditions and the other one is taken as the conclusion, the composition can be formed______ A correct proposition
- 20. Three inequalities are known: (1) AB > 0; (2) C / a > - D / B; (3) BC > ad If two of them are taken as conditions and the remaining one as a conclusion, a correct proposition can be formed Wrong number~ Three inequalities are known: (1) AB > 0; (2) C / AAD