answer:Due to the monotonicity of the logarithmic function: y = log_{2}x is monotonically increasing on the interval (0, +infty) ∴ log_{2}a > log_{2}b Leftrightarrow a > b > 0 Therefore, the correct choice is boxed{C}.

answer:To prove the statement, we consider the possible values of the squares of integers modulo 3. We analyze the remainders when an integer n is squared and then reduced modulo 3. 1. **Consider the possible values of an integer modulo 3:** Every integer n can be expressed as one of the three possible forms modulo 3: [ n equiv 0, 1, text{ or } 2 pmod{3}. ] 2. **Calculate the squares of these values modulo 3:** - If ( n equiv 0 pmod{3} ): [ n^2 equiv 0^2 = 0 pmod{3}. ] - If ( n equiv 1 pmod{3} ): [ n^2 equiv 1^2 = 1 pmod{3}. ] - If ( n equiv 2 pmod{3} ): [ n^2 equiv 2^2 = 4 equiv 1 pmod{3}. ] 3. **Analyze the results:** From the above calculations, we see: [ n^2 pmod{3} text{ can either be } 0 text{ or } 1. ] 4. **Consider the sum of the squares of two integers:** Let the two integers be ( a ) and ( b ). We need to analyze ( a^2 + b^2 ) modulo 3. - We know from step 3 that ( a^2 equiv 0 text{ or } 1 pmod{3} ) and ( b^2 equiv 0 text{ or } 1 pmod{3} ). - The possible values of ( a^2 + b^2 ) modulo 3 are: [ begin{cases} 0 + 0 equiv 0 pmod{3}, 0 + 1 equiv 1 pmod{3}, 1 + 0 equiv 1 pmod{3}, 1 + 1 equiv 2 pmod{3}. end{cases} ] 5. **Condition for divisibility by 3:** For ( a^2 + b^2 ) to be divisible by 3, it must be congruent to 0 modulo 3: [ a^2 + b^2 equiv 0 pmod{3}. ] - From the possible values derived in step 4, we see that ( a^2 + b^2 equiv 0 pmod{3} ) if and only if: [ a^2 equiv 0 pmod{3} text{ and } b^2 equiv 0 pmod{3}. ] - This implies both ( a ) and ( b ) must be congruent to 0 modulo 3 (since only ( n^2 equiv 0 pmod{3} ) when ( n equiv 0 pmod{3} )). # Conclusion: If the sum of the squares of two integers is divisible by 3, then each of these integers must be divisible by 3. [ boxed{} ]

answer:C (Using the elimination method) When a = 0, the original equation has one negative real root, which eliminates options A and D; when a = 1, the original equation has two equal negative real roots, which eliminates option B. Therefore, the correct answer is boxed{text{C}}.

answer:To calculate 5sqrt{3}+(sqrt{4}+2sqrt{3}), we follow these steps: First, we simplify the expression inside the parentheses and outside: 5sqrt{3}+(sqrt{4}+2sqrt{3}) We know that sqrt{4} = 2, so we substitute that in: = 5sqrt{3} + 2 + 2sqrt{3} Now, we combine like terms, 5sqrt{3} and 2sqrt{3}: = (5+2)sqrt{3} + 2 = 7sqrt{3} + 2 Therefore, the final answer is: boxed{7sqrt{3} + 2}