1)Tính nhanh giá trị biểu thức
A=\(\frac{1}{3}-\frac{3}{4}-\left(\frac{-3}{5}\right)+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
B=\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{131}-\left(\frac{-2}{7}\right)+\frac{4}{35}-\frac{7}{18}\)
Bài 1: Tìm x:
a) x - \(\frac{4}{5}=\frac{7}{10}-\frac{3}{4}\)
b)\(2\frac{1}{3}-x=\frac{-5}{9}+2x\)
c) \(\frac{x+3}{2016}+\frac{x+2}{2017}=\frac{x+1}{2018}+\frac{x}{2019}\)
d) \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\)
Bài 1 : Tính nhanh:
a) \(\left(\frac{5}{19}-\frac{1}{511}+\frac{7}{12}\right)-\left(-\frac{1}{511}-\frac{1}{2}+\frac{5}{19}\right)\)
b) \(-\left(\frac{13}{25}-\frac{4}{191}+\frac{2}{51}\right)+\left(\frac{4}{191}+\frac{2}{51}+\frac{3}{5}\right)\)
c) \(\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(A=\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5+\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
Tính: a ) \(-3\frac{3}{4}+\frac{-10}{25}+\frac{-6}{12}\)
b) \(-0,6-\frac{-4}{9}-\frac{16}{15}\)
c ) \(\frac{7}{12}-\left(-\frac{1}{15}\right)-\frac{5}{6}+\frac{2}{3}+\left(-\frac{1}{5}\right)\)
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2022}\right)\)
B=\(\frac{1}{3}-\frac{3}{4}-\left(-0,6\right)+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
C=\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{13}{15}+\frac{11}{15}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+0,6-\frac{1}{3}\)
D=\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}.....-\frac{1}{3.2}-\frac{1}{2.1}\)
C/m các đẳng thức sau:
a) \(\dfrac{1}{a\left(a+1\right)}=\dfrac{1}{a}-\dfrac{1}{a+1}\)
b) \(\dfrac{2}{a\left(a+1\right)\left(a+2\right)}=\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}\)
Tính B=\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...\left(1-\dfrac{1}{n+1}\right)\) với \(n\in N\)
