\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\\ =\dfrac{1}{1}-\dfrac{1}{2020}=\dfrac{2019}{2020}\)
A=1-1/2+1/2-1/3+...+1/2019-1/2020
A=1-1/2020
A=2019/2020
so sánh A và B biết:
A=\(\dfrac{2^{2018}}{2^{2018}+3^{2019}}\)+\(\dfrac{3^{2019}}{3^{2019}+5^{2020}}\)+\(\dfrac{5^{2020}}{5^{2020}+2^{2018}}\)
B=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{5.6}\)+...+\(\dfrac{1}{2019.2020}\).
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{19.20}\)
3. tính:
A=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{49.50}\)
Tính:
a) A = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) +...+ \(\dfrac{1}{998.999}\) + \(\dfrac{1}{999.1000}\)
b) B = \(\dfrac{1}{1.6}\) + \(\dfrac{1}{6.11}\) + \(\dfrac{1}{11.16}\) +...+ \(\dfrac{1}{495.500}\)
c) C = \(\dfrac{1}{1.2.3}\) + \(\dfrac{1}{2.3.4}\) + \(\dfrac{1}{3.4.5}\) +...+ \(\dfrac{1}{998.999.1000}\)
(Mong mn giúp ạ)
tìm số A
A=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{49.50}\)
bạn hãy làm chi tiết
Bài 2: Tìm \(x\) biết:
\(x\)\(\times\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)=1\)
Bài 2: Tìm \(x\) biết:
\(x\)\(\times\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)=1\)
Bài 1 : Tính tổng sau
a) \(A=\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\)
b) \(B=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+....+\dfrac{1}{23.24}+\dfrac{1}{24.25}\)
c) \(C=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
Bài 1 : Tìm 2 số biết hiệu của chúng bằng 5 và 50% số lớn = 1 nửa số bé.
Bài 2 : tính giá trị biểu thức: A = \(\dfrac{9}{1.2}+\dfrac{9}{2.3}+\dfrac{9}{3.4}+...+\dfrac{9}{98.99}+\dfrac{9}{99.100}\)
Bài 3 : Tìm x
a , \(\dfrac{x}{3}-\dfrac{1}{8}=\dfrac{5}{8}\)
