Các câu hỏi tương tự
Chứng minh rằng: B = \(\frac{1.98+2.97+3.96+...+96.3+97.2+98.1}{1.2+2.3+3.4+...+96.97+97.98+98.99}=\frac{1}{2}\)
Chứng minh rằng \(\frac{1.98+2.97+3.96+...+96.3+97.2+98.1}{1.2+2.3+3.4+...+96.97+97.98+98.99}=\frac{1}{2}\)
Chứng minh rằng \(\frac{1.98+2.97+3.96+...+96.3+97.2+98.1}{1.2+2.3+3.4+...+96.97+97.98+98.99}\)=1/2
Tính \(A=\frac{1.98+2.97+3.96+...+97.2+98.1}{1.2+2.3+3.4+...+97.98+98.99}\)
\(S=\frac{1.98+2.97+3.96+...+98.1}{1.2+2.3+3.4+...+98.99}\)
Tính
ngồi rảnh đăng toán cho các bạn làm :D ==
Tính
A=\(\frac{1.98+2.97+3.96+...+98.1}{1.2+2.3+3.4+...+98.99}\)
B=\(\frac{1}{5}+\frac{1}{45}+\frac{1}{117}+\frac{1}{221}+\frac{1}{367}+...< \frac{1}{4}\)
Tính:
B=\(\frac{1.98+2.97+...+98.1}{1.2+2.3+...+98.99}\)
Tính:
B=\(\frac{1.98+2.97+...+98.1}{1.2+2.3+...+98.99}\)
Tính: a) Adfrac{1}{2}+dfrac{1}{2^2}+dfrac{1}{2^3}+...+dfrac{1}{2^{100}}
b) dfrac{1}{1.2}+dfrac{1}{2.3}+dfrac{1}{3.4}+...+dfrac{1}{2023.2024}
cứu tôi mng owiiii :((
Đọc tiếp
Tính: a) A=\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+...+\(\dfrac{1}{2^{100}}\)
b) \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{2023.2024}\)
cứu tôi mng owiiii :((