{ 1/3.5+1/5.7+1/7.9+.....+1/19.21
LN
Những câu hỏi liên quan
Tính A=2/1.3-4/3.5+6/5.7-8/7.9+...-20/19.21
tính 6/3.5+6/5.7+6/7.9+...+6/19.21
MONG CÁC BẠN GIÚP ĐỠ
\(\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+...+\frac{6}{19.21}\)
\(=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{21}\right)\)
\(=3.\frac{2}{7}\)
\(=\frac{6}{7}\)
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các bạn cho mk hỏi câu này
2/3.5+2/5.7+2/7.9+...+2/97.99
thì mk sẽ viết thành
1/3.5+1/5.7+1/7.9+...+1/97.99
hay
2.(1/3.5+1/5.7+1/7.9+...+1/97.99)
giúp mk với
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
~ Hok tốt ~
\(\)
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Viết thành 2 . (1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/97.99
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S=1/1.3+1/3.5+1/5.7+....+1/19.21+1/21
1.Tính hợp lí
a/ 2/3.5 + 2/5.7 + 2/7.9 +...+2/97.99
b/ 1/3.5 + 1/5.7 + 1/7.9 +...+1/97.99
c/1/18 + 1/54 + 1/108 +...+1/990
2.Chứng minh rằng: 1/14 + 1/42 + 1/43 +...+1/79 + 1/80 > 7.12
1/3.5 + 1/5.7 + 1/7.9 + ... + 1/97.99
= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)
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= \(\dfrac{\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}}{2}\)
= \(\dfrac{\dfrac{1}{3}-\dfrac{1}{99}}{2}=\dfrac{16}{99}\)
Bài kia mk nhầm sorry
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\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{19.21}\)
Đặt tên bthuc là A
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{19.21}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\)
\(2A=1-\frac{1}{21}=\frac{20}{21}\)
=>\(A=\frac{20}{21}:2=\frac{10}{21}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{17.19}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{17}-\frac{1}{19}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{19}\right)=\frac{1}{2}.\left(\frac{18}{19}\right)\)
\(=\frac{9}{19}\)
sr nhìn nhầm đề bài
Xem thêm câu trả lời
Tìm \(x\) :
\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\right)\)x \(x\) = \(\frac{9}{7}\)
1/3.5+1/5.7+1/7.9+...+1/97.99 = ?
\(\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{97\times99}\)
\(=\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)+...+\left(\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{99-3}{297}\)
\(=\frac{96}{297}=\frac{32}{99}\)
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