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發問:
1.Solve the inequalities 2x^2+6x+7≧0 or 4x^2+4x+1≦0 2.Given 2x^2+6x+9≡a(x+b)^2+c,where a,b and c are real constants. a)Find the values of a,b and c. b)Using a),or otherwise,write down the range of possibe values of (2/2x^2+6x+9)
最佳解答:
1. Solve the inequalities 2x2 ＋ 6x ＋ 7 ≧ 0 or 4x2 ＋ 4x ＋ 1 ≦ 0 Sol: 2x2 ＋ 6x ＋ 7 ≧ 0 2 ( x2 ＋ 3x ＋ 9/4 ) ＋ 5/2 ≧ 0 2 ( x ＋ 3/2 )2 ＋ 5/2 ≧ 0 for any real number of x, 2x2 ＋ 6x ＋ 7 is always ≧ 0 - ∞ ＜ x ＜ ∞ ....... ( 1 ) 4x2 ＋ 4x ＋ 1 ≦ 0 ( 2x ＋ 1 )2 ≦ 0 x ＝ - 1/2 ....... ( 2 ) combine ( 1 ) and ( 2 ) - ∞ ＜ x ＜ ∞ Ans: - ∞ ＜ x ＜ ∞ 2.Given 2x2 ＋ 6x ＋ 9 ≡ a ( x ＋ b )2 ＋ c, where a, b and c are real constants. ( a ) Find the values of a, b and c. ( b ) Using ( a ),or otherwise, write down the range of possible values of 2/(2x2+6x+9) Sol: ( a ) 2x2 ＋ 6x ＋ 9 ≡ a ( x ＋ b )2 ＋ c 2 ( x2 ＋ 3x ＋ 9/4 ) ＋ 9/2 ≡ a ( x ＋ b )2 ＋ c 2 ( x ＋ 3/2 )2 ＋ 9/2 ≡ a ( x ＋ b )2 ＋ c a ＝ 2, b ＝ 3/2, c ＝ 9/2 ( b ) 2 ( x ＋ 3/2 )2 ＋ 9/2 ≧ 9/2 The maximum value of 2 ( x ＋ 3/2 )2 ＋ 9/2 is ∞ and the minimum value of 2 ( x ＋ 3/2 )2 ＋ 9/2 is 9/2 therefore ∞ ＞ 2x2 ＋ 6x ＋ 9 ≧ 9/2 1/∞ ＜ 1/(2x2+6x+9) ≦ 1/(9/2) 1/∞ ＜ 1/(2x2+6x+9) ≦ 2/9 1/∞ is very close to 0 so, 1/∞ ＜ 1/(2x2+6x+9) ≦ 2/9 0 ＜ 1/(2x2+6x+9) ≦ 2/9 0 ＜ 2/(2x2+6x+9) ≦ 4/9 ( you can write it down as (0,9/4] ) Ans: ( a ) a ＝ 2, b ＝ 3/2, c ＝ 9/2 ( b ) 0 ＜ 2/(2x2+6x+9) ≦ 4/9
其他解答:
1. 2x2 ＋ 6x ＋ 7 ≧ 0 2 ( x2 ＋ 3x ＋ 9/4 ) ＋ 5/2 ≧ 0 2 ( x ＋ 3/2 )2 ＋ 5/2 ≧ 0 for any real number of x, 2x2 ＋ 6x ＋ 7 is always ≧ 0 - ∞ ＜ x ＜ ∞ ....... ( 1 ) 4x2 ＋ 4x ＋ 1 ≦ 0 ( 2x ＋ 1 )2 ≦ 0 x ＝ - 1/2 ....... ( 2 ) combine ( 1 ) and ( 2 ) so, - ∞ ＜ x ＜ ∞ 2. ( a ) 2x2 ＋ 6x ＋ 9 ≡ a ( x ＋ b )2 ＋ c 2 ( x2 ＋ 3x ＋ 9/4 ) ＋ 9/2 ≡ a ( x ＋ b )2 ＋ c 2 ( x ＋ 3/2 )2 ＋ 9/2 ≡ a ( x ＋ b )2 ＋ c so, a ＝ 2, b ＝ 3/2, c ＝ 9/2 ( b ) 2 ( x ＋ 3/2 )2 ＋ 9/2 ≧ 9/2 The maximum value of 2 ( x ＋ 3/2 )2 ＋ 9/2 is ∞ and the minimum value of 2 ( x ＋ 3/2 )2 ＋ 9/2 is 9/2 therefore ∞ ＞ 2x2 ＋ 6x ＋ 9 ≧ 9/2 1/∞ ＜ 1/(2x2+6x+9) ≦ 1/(9/2) 1/∞ ＜ 1/(2x2+6x+9) ≦ 2/9 1/∞ is very close to 0 so, 1/∞ ＜ 1/(2x2+6x+9) ≦ 2/9 0 ＜ 1/(2x2+6x+9) ≦ 2/9 0 ＜ 2/(2x2+6x+9) ≦ 4/9
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A.Math發問:
1.Solve the inequalities 2x^2+6x+7≧0 or 4x^2+4x+1≦0 2.Given 2x^2+6x+9≡a(x+b)^2+c,where a,b and c are real constants. a)Find the values of a,b and c. b)Using a),or otherwise,write down the range of possibe values of (2/2x^2+6x+9)
最佳解答:
1. Solve the inequalities 2x2 ＋ 6x ＋ 7 ≧ 0 or 4x2 ＋ 4x ＋ 1 ≦ 0 Sol: 2x2 ＋ 6x ＋ 7 ≧ 0 2 ( x2 ＋ 3x ＋ 9/4 ) ＋ 5/2 ≧ 0 2 ( x ＋ 3/2 )2 ＋ 5/2 ≧ 0 for any real number of x, 2x2 ＋ 6x ＋ 7 is always ≧ 0 - ∞ ＜ x ＜ ∞ ....... ( 1 ) 4x2 ＋ 4x ＋ 1 ≦ 0 ( 2x ＋ 1 )2 ≦ 0 x ＝ - 1/2 ....... ( 2 ) combine ( 1 ) and ( 2 ) - ∞ ＜ x ＜ ∞ Ans: - ∞ ＜ x ＜ ∞ 2.Given 2x2 ＋ 6x ＋ 9 ≡ a ( x ＋ b )2 ＋ c, where a, b and c are real constants. ( a ) Find the values of a, b and c. ( b ) Using ( a ),or otherwise, write down the range of possible values of 2/(2x2+6x+9) Sol: ( a ) 2x2 ＋ 6x ＋ 9 ≡ a ( x ＋ b )2 ＋ c 2 ( x2 ＋ 3x ＋ 9/4 ) ＋ 9/2 ≡ a ( x ＋ b )2 ＋ c 2 ( x ＋ 3/2 )2 ＋ 9/2 ≡ a ( x ＋ b )2 ＋ c a ＝ 2, b ＝ 3/2, c ＝ 9/2 ( b ) 2 ( x ＋ 3/2 )2 ＋ 9/2 ≧ 9/2 The maximum value of 2 ( x ＋ 3/2 )2 ＋ 9/2 is ∞ and the minimum value of 2 ( x ＋ 3/2 )2 ＋ 9/2 is 9/2 therefore ∞ ＞ 2x2 ＋ 6x ＋ 9 ≧ 9/2 1/∞ ＜ 1/(2x2+6x+9) ≦ 1/(9/2) 1/∞ ＜ 1/(2x2+6x+9) ≦ 2/9 1/∞ is very close to 0 so, 1/∞ ＜ 1/(2x2+6x+9) ≦ 2/9 0 ＜ 1/(2x2+6x+9) ≦ 2/9 0 ＜ 2/(2x2+6x+9) ≦ 4/9 ( you can write it down as (0,9/4] ) Ans: ( a ) a ＝ 2, b ＝ 3/2, c ＝ 9/2 ( b ) 0 ＜ 2/(2x2+6x+9) ≦ 4/9
其他解答:
1. 2x2 ＋ 6x ＋ 7 ≧ 0 2 ( x2 ＋ 3x ＋ 9/4 ) ＋ 5/2 ≧ 0 2 ( x ＋ 3/2 )2 ＋ 5/2 ≧ 0 for any real number of x, 2x2 ＋ 6x ＋ 7 is always ≧ 0 - ∞ ＜ x ＜ ∞ ....... ( 1 ) 4x2 ＋ 4x ＋ 1 ≦ 0 ( 2x ＋ 1 )2 ≦ 0 x ＝ - 1/2 ....... ( 2 ) combine ( 1 ) and ( 2 ) so, - ∞ ＜ x ＜ ∞ 2. ( a ) 2x2 ＋ 6x ＋ 9 ≡ a ( x ＋ b )2 ＋ c 2 ( x2 ＋ 3x ＋ 9/4 ) ＋ 9/2 ≡ a ( x ＋ b )2 ＋ c 2 ( x ＋ 3/2 )2 ＋ 9/2 ≡ a ( x ＋ b )2 ＋ c so, a ＝ 2, b ＝ 3/2, c ＝ 9/2 ( b ) 2 ( x ＋ 3/2 )2 ＋ 9/2 ≧ 9/2 The maximum value of 2 ( x ＋ 3/2 )2 ＋ 9/2 is ∞ and the minimum value of 2 ( x ＋ 3/2 )2 ＋ 9/2 is 9/2 therefore ∞ ＞ 2x2 ＋ 6x ＋ 9 ≧ 9/2 1/∞ ＜ 1/(2x2+6x+9) ≦ 1/(9/2) 1/∞ ＜ 1/(2x2+6x+9) ≦ 2/9 1/∞ is very close to 0 so, 1/∞ ＜ 1/(2x2+6x+9) ≦ 2/9 0 ＜ 1/(2x2+6x+9) ≦ 2/9 0 ＜ 2/(2x2+6x+9) ≦ 4/9
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