answer:First, we solve for the roots of the equation. Then, based on the relationship between the sides of the triangle, we determine that when x = 4, the three line segments of lengths 4, 4, and 8 cannot form a triangle. Therefore, the length of the legs of the isosceles triangle is boxed{5}. This problem tests our understanding of triangles from the perspective of their sides, involving the method of classification and discussion. When finding the length of a triangle's side, we should not make hasty judgments. Instead, we should cultivate the habit of verifying whether the three side lengths can form a triangle, and discard those that do not meet the requirements of the problem.

answer:To convert the decimal number 0.32 into a vulgar fraction, we can write it as 32/100 since 0.32 means 32 hundredths. However, we want to simplify this fraction to have a numerator of 16. First, we simplify 32/100 by dividing both the numerator and the denominator by their greatest common divisor (GCD), which is 4: 32 ÷ 4 = 8 100 ÷ 4 = 25 So, 32/100 simplifies to 8/25. Now, we want the numerator to be 16, which is twice the simplified numerator of 8. To keep the fraction equivalent, we must also multiply the denominator by the same factor: 8 × 2 = 16 25 × 2 = 50 Therefore, the vulgar fraction with a numerator of 16 and equivalent to the decimal 0.32 is boxed{16/50} .

answer:To prove dfrac{1}{a+b}+dfrac{1}{b+c}=dfrac{3}{a+b+c}, it suffices to show that (b+c)(a+b+c)+(a+b)(a+b+c)=3(a+b)(b+c), which is equivalent to proving a^{2}-b^{2}+c^{2}-ac=0 (1). In triangle ABC, since the degrees of angles A, B, C form an arithmetic sequence, we have B=60^{circ}. Therefore, cos B= dfrac{a^{2}+c^{2}-b^{2}}{2ac}, which implies a^{2}-b^{2}+c^{2}-ac=0, thus equation (1) is evidently true, completing the proof. boxed{}

answer:# Detailed Solution: To solve this problem, we need to find all functions ( f : mathbb{N} to mathbb{N} ) such that ( forall x in mathbb{N}, quad f(f(x)) = x + 1 ). We'll calculate ( f(f(f(x))) ) in two different ways to derive more properties about the function ( f ). 1. **First Calculation of ( f(f(f(x))) ):** Given the initial functional equation ( f(f(x)) = x + 1 ), we apply ( f ) to both sides: [ f(f(f(x))) = f(x + 1) ] This follows directly from the substitution of ( f(f(x)) = x + 1 ) into the function ( f ). 2. **Second Calculation of ( f(f(f(x))) ):** Let ( y = f(x) ). Then, according to the initial equation: [ f(f(y)) = y + 1 ] By substituting ( y = f(x) ) back into this equation, we get: [ f(f(f(x))) = f(y) + 1 = f(f(x)) + 1 ] Since ( f(f(x)) = x + 1 ), this becomes: [ f(f(f(x))) = f(x) + 1 ] 3. **Equating Both Results:** We now have two expressions for ( f(f(f(x))) ): [ f(f(f(x))) = f(x + 1) ] and [ f(f(f(x))) = f(x) + 1 ] Thus, [ f(x + 1) = f(x) + 1 ] 4. **Solving the Recurrence Relation:** The equation ( f(x + 1) = f(x) + 1 ) is a simple recurrence relation. Solving it via induction: - Base case: Assume ( f(0) = c ). - Inductive step: Assume ( f(k) = k + c ) holds for some ( k ). Then: [ f(k + 1) = f(k) + 1 = k + c + 1 = (k + 1) + c ] Using mathematical induction, we confirm that: [ f(x) = x + c quad text{for all} x in mathbb{N} ] 5. **Substituting Back to Check Validity:** Substitute ( f(x) = x + c ) back into the original equation: [ f(f(x)) = f(x + c) = (x + c) + c = x + 2c ] The original functional equation requires ( f(f(x)) = x + 1 ). Thus, we must have: [ x + 2c = x + 1 implies 2c = 1 implies c = frac{1}{2} ] Since ( f ) maps ( mathbb{N} ) to ( mathbb{N} ), ( c ) must be a natural number. However, ( frac{1}{2} ) is not a natural number. Therefore, the supposed function ( f(x) = x + frac{1}{2} ) cannot be a valid solution. # Conclusion: There are no functions ( f: mathbb{N} to mathbb{N} ) satisfying ( forall x in mathbb{N}, quad f(f(x)) = x + 1 ): [ boxed{text{No solutions}} ]