A = \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
2A = 2 . \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
2A = \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
2A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
2A = \(\frac{1}{3}-\frac{1}{99}\)
2A = \(\frac{32}{99}\)
A = \(\frac{32}{99}\div2\)
A =\(\frac{16}{99}\)
_HT_
a) Tính tổng : \(S=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{97.99}\)
tính:a/\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+......+\frac{2}{97.99}\)
Bài 1:Tính nhanh. \(\frac{4}{3.5}+\frac{4}{5.7}+....+\frac{4}{97.99}\)
Bài 2: Tìm x. a) \(2.4^x-18=110\)
b) \(\left(\frac{3}{2}.x-1\right)^5=1\)
Tìm: \(\frac{x-2}{3}+\frac{x-2}{3.5}+\frac{x-2}{5.7}+...+\frac{x-2}{97.99}=\frac{-49}{99}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{97.99}\)
Câu 1:Rút gọn các biểu thức:
A=\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}-\frac{5}{4}.\frac{13}{99}+\frac{5}{99}.\frac{1}{4}\)
Câu 2: So sánh:
A=\(\frac{2013}{2014}+\frac{2016}{2015}\)và \(\frac{2014}{2015}+\frac{2017}{2016}\)
Câu 3: Cho f(x)=ax2+bx+c. Biết 7a+b=0. Chứng minh rằng: f(10).f(-3)\(\ge\)0
Tính \(\frac{1}{3.1}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+...+\frac{1}{97.99}-\frac{1}{98.100}\)
a,tính \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{19.21}\)
b,CMR A\(=\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{(2n-1).(2n+1)}\le\frac{1}{2}\)
\(B=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.6\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(C=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)
\(D=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)