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If the equation (a-6) x2-8x + 6 = 0 of X has real roots, then the maximum value of integer a is () A. 6B. 7C. 8D. 9
When a-6 = 0, i.e. a = 6, the equation is - 8x + 6 = 0, the solution is x = 68 = 34; when a-6 ≠ 0, i.e. a ≠ 6, △ = (- 8) 2-4 (a-6) × 6 = 208-24a ≥ 0, solve the above equation, a ≤ 263 ≈ 8.6, take the largest integer, i.e. a = 8
If the equation (a-6) x2-8x + 6 = 0 of X has real roots, then the maximum value of integer a is ()
A. 6B. 7C. 8D. 9
When a-6 = 0, i.e. a = 6, the equation is - 8x + 6 = 0, the solution is x = 68 = 34; when a-6 ≠ 0, i.e. a ≠ 6, △ = (- 8) 2-4 (a-6) × 6 = 208-24a ≥ 0, solve the above equation, a ≤ 263 ≈ 8.6, take the largest integer, i.e. a = 8
Given that real numbers a and B satisfy a & # 178; – 6A + 4 = 0, B & # 178; – 6B + 4 = 0, and a ≠ B, then what is the value of B / A + A / b? Can you help me? I've been struggling all afternoon!
According to the meaning of the problem, a and B are the two real roots of the equation x ^ 2-6x + 4 = 0
So (according to Veda's theorem)
a+b=6
ab=4
b/a+a/b
=(a^2+b^2)/(ab)
=(a^2+b^2+2ab)/(ab)-2
=(a+b)^2/(ab)-2
=6^2/4-2
=9-2
=7
Given that the real numbers a and B satisfy a & # 178; - 6A + 4 = 0, B & # 178; - 6B + 4 = 0 respectively, what is the value of B / A + A / b? (solved)
Given that real numbers a and B satisfy a2-6a + 4 = 0 and b2-6b + 4 = 0 respectively, and a ≠ B, then the value of Ba + AB is ()
A. 7B. -7C. 11D. -11
According to the meaning of the question: A and B are two parts of the equation x2-6x + 4 = 0, a + B = 6, ab = 4, then the original formula = (a + b) 2 − 2abab = 36 − 84 = 7
If x = A and y = B are solutions of equation 2x + y = 3, then 20-6a-3b=
If x = A and y = B are solutions of equation 2x + y = 3, then 20-6a-3b=
2x+y=3
2a+b=3
6a+3b=9
∴20-6a-3b=20-(6a+3b)=20-9=11
If x = ay = B is a solution of the equation 2x + y = 1, then 6A + 3B + 2=______ ．
Substituting x = ay = B into the equation 2x + y = 1, we get: 2A + B = 1, then 6A + 3B + 2 = 3 (2a + b) + 2 = 3 × 1 + 2 = 5, so the answer is: 5
If x = ay = B is a solution of the equation 2x + y = 1, then 6A + 3B + 2=______ ．
Substituting x = ay = B into the equation 2x + y = 1, we get: 2A + B = 1, then 6A + 3B + 2 = 3 (2a + b) + 2 = 3 × 1 + 2 = 5, so the answer is: 5
If x = ay = B is the solution of the equation 2x + y = 0, then 6A + 3B + 2=______ ．
Substituting x = ay = B into the equation 2x + y = 0, we get 2A + B = 0, ∩ 6A + 3B + 2 = 3 (2a + b) + 2 = 2
If the nonempty set a = {X / 2A + 1 ≤ x ≤ 3a-5}, B = {X / 13 ≤ x ≤ 22}, find the set that a contains all the real numbers a that (a ∩ b) holds
2a+1≤3a－5,
13≤2a+1,
3a－5≤22,
That is, 6 ≤ a ≤ 9, a = {A / 6 ≤ a ≤ 9},
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