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Tính A=\(\frac{\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{101\cdot400}}{\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+...+\frac{1}{299\cdot400}}\)
Tính \(\frac{A}{B}\), biết rằng :
\(A=\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{101\cdot400}\)
\(B=\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+...+\frac{1}{299\cdot400}\)
Tính \(\frac{A}{B}\), biết rằng :
\(A=\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{101\cdot400}\)
\(B=\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+...+\frac{1}{299\cdot400}\)
1> Tính A/B
A= \(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+.....+\frac{1}{101\cdot400}\)
B = \(\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+.....+\frac{1}{299\cdot400}\)
2> Cho a,b thộc N*
CMR :\(\frac{a}{b}+\frac{b}{a}\)lớn hơn hoặc bằng 2
nhanh nha đang cần gấp
thanks so much
a) Rút gọn:frac{frac{1}{1cdot300}+frac{1}{2cdot301}+frac{1}{3cdot302}+...+frac{1}{101cdot400}}{frac{1}{1cdot102}+frac{1}{2cdot103}+frac{1}{3cdot104}+...+frac{1}{299cdot400}}b) CMR: 1cdot3cdot5cdot7cdot9cdot...cdot197cdot199 frac{101}{2}cdotfrac{102}{2}cdotfrac{103}{2}cdot...cdotfrac{200}{2}.c) Cho: Afrac{1}{1cdot2}+frac{1}{2cdot3}+frac{1}{3cdot4}+...+frac{1}{199cdot200}. Bfrac{1}{101cdot200}+frac{1}{102cdot199}+...+frac{1}{199cdot102}+frac{1}{200cdot101}.d) Tìm số tự nhiên n lớn nhất có ba ch...
Đọc tiếp
a) Rút gọn:
\(\frac{\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{101\cdot400}}{\frac{1}{1\cdot102}+\frac{1}{2\cdot103}+\frac{1}{3\cdot104}+...+\frac{1}{299\cdot400}}\)
b) CMR: \(1\cdot3\cdot5\cdot7\cdot9\cdot...\cdot197\cdot199\)= \(\frac{101}{2}\cdot\frac{102}{2}\cdot\frac{103}{2}\cdot...\cdot\frac{200}{2}\).
c) Cho: A=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{199\cdot200}\).
B=\(\frac{1}{101\cdot200}+\frac{1}{102\cdot199}+...+\frac{1}{199\cdot102}+\frac{1}{200\cdot101}\).
d) Tìm số tự nhiên n lớn nhất có ba chữ số sao cho n chia 8 dư 7,chia 31 dư 28.
e) Tìm số nguyên tố \(\overline{ab}\) (a>0>b),sao cho \(\overline{ab}-\overline{ba}\)là số chính phương.
\(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{299\cdot400}\)chứng minh rằng \(\frac{1}{299}\left(\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right)\)=\(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...+\frac{1}{299\cdot400}\)
\(\frac{1}{1\cdot300}+\frac{1}{2\cdot301}+\frac{1}{3\cdot302}+...\frac{1}{101\cdot400}\)=
E=\(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+\frac{1}{3\cdot103}+...+\frac{1}{10\cdot110}\)và F=\(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+\frac{1}{3\cdot13}+...+\frac{1}{100\cdot110}\)
tính tỉ số\(\frac{E}{F}\)
a)A=\(\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+\frac{1}{5\cdot7\cdot9}+...+\frac{1}{25\cdot27\cdot29}\)
b)\(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)\cdot x=\frac{1}{1.11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)