answer:The direction vector of the first line is (begin{pmatrix} 1 4 5 end{pmatrix}) and the direction vector of the second line is (begin{pmatrix} 3 1 2 end{pmatrix}). Since these vectors are not the same scalar multiple, the two lines are not parallel. For the lines to be skew, they must also not intersect. We set up the equations based on equaling the point formulations: [ begin{pmatrix} 2 3 a end{pmatrix} + t begin{pmatrix} 1 4 5 end{pmatrix} = begin{pmatrix} 7 0 1 end{pmatrix} + u begin{pmatrix} 3 1 2 end{pmatrix} ] Breaking it down: 1. (2 + t = 7 + 3u) 2. (3 + 4t = 2u) 3. (a + 5t = 1 + 2u) From (1), (t = 7 + 3u - 2), simplify to (t = 3u + 5). Substitute (t) into (2), [ 3 + 4(3u + 5) = 2u ] [ 3 + 12u + 20 = 2u ] [ 12u - 2u = -23 ] [ 10u = -23 ] [ u = -frac{23}{10} ] Insert (u) into (t = 3u + 5): [ t = 3left(-frac{23}{10}right) + 5 = -frac{69}{10} + 5 = -frac{69}{10} + frac{50}{10} = -frac{19}{10} ] Now, check if the third component agrees: [ a + 5left(-frac{19}{10}right) = 1 + 2left(-frac{23}{10}right) ] [ a - frac{95}{10} = 1 - frac{46}{10} ] [ a = 1 - frac{46}{10} + frac{95}{10} = frac{50}{10} = 5 ] The two lines intersect when (a = 5). Therefore, they are skew for all (a neq 5), or (a in boxed{(-infty, 5) cup (5, infty)}).

answer:Given: ( M_n ) is the least common multiple (LCM) of the numbers ( 1, 2, 3, ldots, n ). For example, ( M_1 = 1 ), ( M_2 = 2 ), ( M_3 = 6 ), ( M_4 = 12 ), ( M_5 = 60 ), ( M_6 = 60 ). We need to determine for which integer ( n ) it holds that ( M_{n-1} = M_n ). 1. **First Case: ( n ) is a Power of a Prime Number** - If ( n ) is a power of a prime number, say ( n = p^k ) where ( p ) is a prime, then ( M_n = text{lcm}(1, 2, ldots, n) = text{lcm}(M_{n-1}, n) = text{lcm}(M_{n-1}, p^k) ). - Since ( M_{n-1} ) already includes ( p ) raised to the highest power less than ( p^k ), it means ( M_n = p cdot M_{n-1} ). - Therefore, ( M_{n-1} neq M_n ). 2. **Second Case: ( n ) is Not a Power of a Prime Number** - If ( n ) is not a power of a prime number, then ( n ) can be expressed as ( n = ab ) where ( 1 < a < n ) and ( 1 < b < n ), and there is no prime number that divides ( n ) exactly. - Therefore, ( a leq n-1 ) and ( b leq n-1 ). - Considering ( a ) and ( b ) are co-prime, i.e., ((a, b) = 1), it follows that ( a | M_{n-1} ) and ( b | M_{n-1} ). - Therefore, ( n = ab mid M_{n-1} ). Since ( n ) divides ( M_{n-1} ), we have: [ M_n = M_{n-1} ] **Conclusion:** The necessary and sufficient condition for ( M_{n-1} = M_n ) is that ( n ) is not a power of any prime number. (boxed{text{The integer } n text{ should not be a power of any prime number.}})

answer:We apply the Cauchy-Schwarz inequality with the weight vector (sqrt{2}, sqrt{3}, sqrt{1}) and the vector (sqrt{2x+20}, sqrt{26-2x}, sqrt{3x}). This results in: [ (sqrt{2}^2 + sqrt{3}^2 + sqrt{1}^2) left(2(2x + 20) + 3(26 - 2x) + 1 times 3xright) geq left(sqrt{2x + 20} + sqrt{26 - 2x} + sqrt{3x}right)^2 ] Simplifying: [ 8 left(2 times (2x + 20) + 3 times (26 - 2x) + 3xright) geq left(sqrt{2x + 20} + sqrt{26 - 2x} + sqrt{3x}right)^2 ] Calculating the expression inside the parentheses: [ 4(2x + 20) + 3(26 - 2x) + 3x = 8x + 80 + 78 - 6x + 3x = 5x + 158 ] Thus: [ 8(5x + 158) = 40x + 1264 geq left(sqrt{2x + 20} + sqrt{26 - 2x} + sqrt{3x}right)^2 ] Check for equality alignment: When x=10: [ left(sqrt{20 + 20} + sqrt{26 - 20} + sqrt{30}right)^2 = left(2sqrt{10} + sqrt{6} + sqrt{30}right)^2 ] Plugging in x=10 back into the squared form, we get 1264, confirming the maximum is potentially at x=10. Hence, squaring the expression: [ left(sqrt{2x + 20} + sqrt{26 - 2x} + sqrt{3x}right)^2 leq 1264 ] [sqrt{2x + 20} + sqrt{26 - 2x} + sqrt{3x} leq sqrt{1264} = 4sqrt{79}] Conclusion: The maximum value of the expression, when x = 10, is boxed{4sqrt{79}}.

answer:1. **Introduction to Negative Numbers:** To ensure that the subtraction operation is always possible, negative numbers were introduced. By extending the set of natural numbers to include their additive inverses, we can perform subtraction freely. - For example, in purely natural numbers, (5 - 7) would not be defined because (7) is not less than (5). However, introducing negative numbers allows us to write (5 - 7 = -2). 2. **Rational Numbers:** To make division always executable, rational numbers (fractions) were introduced. This allows every division operation, except by zero, to have a result within the number system. - For example, the division (5 div 2) results in ( frac{5}{2}), which is a rational number. Similarly, any integer divided by another non-zero integer results in a rational number. 3. **Irrational Numbers:** To accommodate the operations like extracting roots of numbers that are not perfect powers, irrational numbers were introduced. This generalization allows us to have roots for all positive rational numbers. - For example, the square root of 2 ((sqrt{2})) cannot be expressed as a fraction but is known to exist and is an irrational number. The inclusion of irrational numbers ensures that we can find the square root of any positive rational number. 4. **Imaginary Numbers:** Finally, imaginary numbers were introduced to handle the square roots of negative numbers. This keeps us from being halted by negative values under the square root. - For example, the square root of (-1) is not a real number because no real number squared gives (-1). This led to the introduction of the imaginary unit (i) where (i^2 = -1). The number (sqrt{-1}) is defined as (i), and this extends our number system to include complex numbers (a + bi) where (a) and (b) are real numbers. # Conclusion: The progression from integers to rationals, irrationals, and finally imaginary numbers represents a systematic effort to generalize the concept of numbers to ensure that all basic operations are always executable within the number system. Each stage of this progression resulted in a broader and more encompassing number system: - Negative numbers for subtraction. - Rational numbers for division. - Irrational numbers for extracting roots. - Imaginary numbers for dealing with the square roots of negative numbers. With these extensions, we now have a comprehensive understanding and system that encompasses nearly any numerical operation commonly encountered in mathematics. blacksquare