answer:(1) Since the function f(x) = e^x - ln(x + m), Therefore, f'(x) = e^x - frac{1}{x + m}. Also, since x = 0 is an extremum point of f(x), Therefore, f'(0) = 1 - frac{1}{m} = 0, solving this gives m = 1. (2) From (1), we know that the function f(x) = e^x - ln(x + 1), whose domain is (-1, +infty). Since f'(x) = frac{e^x(x + 1) - 1}{x + 1}. Let g(x) = e^x(x + 1) - 1, Then g'(x) = e^x(x + 1) + e^x > 0, Thus, g(x) is increasing on (-1, +infty), Also, since g(0) = 0, Therefore, when x > 0, g(x) > 0, i.e., f'(x) > 0; When -1 < x < 0, g(x) < 0, f'(x) < 0. Hence, f(x) is decreasing on (-1, 0) and increasing on (0, +infty); Thus, the answer is: m = boxed{1}, the interval of increase is boxed{(0, +infty)}, and the interval of decrease is boxed{(-1, 0)}. (1) Find the derivative of the original function. Since x = 0 is an extremum point of f(x), we find the value of m by setting the derivative at this point to zero. (2) Substitute the value of m into the function's expression, then determine the intervals of monotonicity of the original function by examining where the derivative is greater than or less than zero. This problem examines the use of derivatives to study the monotonicity of functions, comprehensively testing students' ability to analyze and solve problems. Proficiency in the basics of functions and derivatives is key to solving this problem.

answer:Given overrightarrow{a}=(-1,lambda,-2) and overrightarrow{b}=(2,-1,1), with the angle between overrightarrow{a} and overrightarrow{b} being 120^{circ}, we can find the value of lambda by using the formula for the cosine of the angle between two vectors: [ cos theta = frac{overrightarrow{a} cdot overrightarrow{b}}{|overrightarrow{a}| cdot |overrightarrow{b}|} ] Given theta = 120^{circ}, we have: [ cos 120^{circ} = -frac{1}{2} = frac{overrightarrow{a} cdot overrightarrow{b}}{|overrightarrow{a}| cdot |overrightarrow{b}|} ] The dot product overrightarrow{a} cdot overrightarrow{b} is calculated as: [ overrightarrow{a} cdot overrightarrow{b} = (-1) cdot 2 + lambda cdot (-1) + (-2) cdot 1 = -2 - lambda - 2 ] The magnitudes of overrightarrow{a} and overrightarrow{b} are: [ |overrightarrow{a}| = sqrt{(-1)^2 + lambda^2 + (-2)^2} = sqrt{1 + lambda^2 + 4} = sqrt{5 + lambda^2} ] [ |overrightarrow{b}| = sqrt{2^2 + (-1)^2 + 1^2} = sqrt{4 + 1 + 1} = sqrt{6} ] Substituting these into the cosine formula gives: [ -frac{1}{2} = frac{-2 - lambda - 2}{sqrt{5 + lambda^2} cdot sqrt{6}} ] Simplifying, we get: [ -frac{1}{2} = frac{-4 - lambda}{sqrt{5 + lambda^2} cdot sqrt{6}} ] Multiplying both sides by 2sqrt{5 + lambda^2} cdot sqrt{6}: [ -sqrt{5 + lambda^2} cdot sqrt{6} = -8 - 2lambda ] Squaring both sides to eliminate the square root gives a quadratic equation in lambda. Solving this equation yields two possible values for lambda: [ lambda = -1 quad text{or} quad lambda = 17 ] Therefore, the correct answers are: [ boxed{A quad text{and} quad C} ]

answer:Let's denote the certain number as ( x ). According to the given information, we perform the following operations on ( x ): 1. Add 20 to ( x ): ( x + 20 ) 2. Multiply the sum by 2: ( 2(x + 20) ) 3. Divide the product by 2: ( frac{2(x + 20)}{2} ) 4. Subtract 2 from the quotient: ( frac{2(x + 20)}{2} - 2 ) Now, let's simplify the expression: ( frac{2(x + 20)}{2} - 2 = (x + 20) - 2 = x + 18 ) The problem states that the remainder left is a fraction of 88. This means that ( x + 18 ) is a fraction of 88. To find out what fraction it is, we can set up the following equation: ( x + 18 = frac{p}{q} cdot 88 ) where ( frac{p}{q} ) is the fraction we are looking for. Since ( x + 18 ) is a fraction of 88, it must be less than 88. Therefore, ( p < q ) because a proper fraction has a numerator smaller than its denominator. To find the fraction ( frac{p}{q} ), we need more information about the value of ( x ) or the specific fraction that ( x + 18 ) represents of 88. Without additional information, we cannot determine the exact fraction. However, we can express the fraction in terms of ( x ) as follows: ( frac{x + 18}{88} ) This expression represents the fraction of boxed{88} that is left after performing the given operations on the certain number ( x ).

answer:**Analysis** This problem tests the method of inductive reasoning. Observing the right side of each equation, it is found that starting from the third term, each term is the sum of its two preceding terms, from which the conclusion can be drawn. **Solution** Observation reveals that the values of the equations form the sequence 1, 3, 4, 7, 11, (ldots), The pattern is that from the third term onwards, each term equals the sum of its two immediate preceding terms, The value sought is the tenth term in the sequence. Continuing to write out this sequence as 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, (ldots), The tenth term is 123, that is a^{10} + b^{10} = 123, Therefore, the correct choice is boxed{C}.