Give \(B=\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{96}=\dfrac{a}{b}\). Prove that: \(a⋮97\)
Prove that:
\(\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}+...+\dfrac{1}{101}\notin N\)
Find x:
\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\left(x+1\right):2}=1\dfrac{1991}{1993}\)
The sum \(\dfrac{1}{50}+\dfrac{1}{51}+...+\dfrac{1}{99}=\dfrac{a}{b}\). Prove that: \(a⋮149\)
Give \(A=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\),prove that: \(A\notin N\)
Calculate:
\(1^2+2^2+3^2+...+98^2\)
Find the product of 98 first number in this series:
\(1\dfrac{1}{3};1\dfrac{1}{8};1\dfrac{1}{15};1\dfrac{1}{24};...\)
\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}{\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}}\)
\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{37.38.39}\)
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^8}\)
The bottom of a triangle increases 20%, the height decreases 20%. The area:
A. Don't change
B. Increases 4%
C. Decreases 4%
D. Decreases 10%
The side of a cube increases 50%. It's volume increases:
A. 50%
B. 100%
C. 237,5%
D. 125%
The side of a square increases 20%. It's area increases:
A. 144%
B. 40%
C. 80%
D. 44%
\(C=10101.\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3.7.11.13.17}\right)\)
Prove that these fractions can be written by the sum of the fractions that had the numberator 1, the denominator \(\ge0\) and different:
\(\dfrac{1}{6}\);\(\dfrac{15}{22}\) and \(\dfrac{5}{11}\)
Find \(x,y\in Z\):
\(\dfrac{x}{6}-\dfrac{2}{y}=\dfrac{1}{30}\)
\(\dfrac{5}{x}-\dfrac{y}{3}=\dfrac{1}{6}\)
Find \(x,y\in N\):
\(\dfrac{4}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)
\(\dfrac{x}{3}-\dfrac{4}{y}=\dfrac{1}{5}\)