tính
A=\(\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+.....+\frac{6}{49.50}\)
Có lời giải
\(E=\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}-\frac{10}{9.11}+...+\frac{100}{99.101}\)
B=\(\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}-\frac{10}{9.11}+\frac{12}{11.13}-...+\frac{100}{99.101}\)
TINH B
B=2(2/3.5 - 2/ 5.7 +....................+ 2/99.101)
B=2(1/3.5 -2/5.7+..............+1/99.100)
B=2(1/3-1/5+1/5-.............+1/99-1/100)
B=2(1/3-1/100)
B=2.97/100
B=97/50
B=\(\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}-\frac{10}{9.11}+\frac{12}{11.13}-...+\frac{100}{99.101}\)
TINH B
B=\(\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}-\frac{10}{9.11}+\frac{12}{11.13}-...+\frac{100}{99.101}\)
TINH B
Tìm giá trị của biểu thức \(P=\frac{2}{1.3}-\frac{4}{3.5}+\frac{6}{5.7}+\frac{8}{7.9}+...-\frac{96}{95.97}+\frac{98}{97.99}\)
Tinh
F=\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)có lời giải nhé😘😘😘
F = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
F = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
F = \(\frac{1}{3}-\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{1}{7}-\frac{1}{7}\right)-\left(\frac{1}{9}-\frac{1}{9}\right)-...-\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
F = \(\frac{1}{3}-\frac{1}{99}\)
F = \(\frac{32}{99}\)
\(F=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)
\(\Rightarrow F=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{97}-\frac{1}{99}\)
\(\Rightarrow F=\frac{1}{3}-\frac{1}{99}\)
\(\Rightarrow F=\frac{32}{99}\)
\(F=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(F=1.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(F=1.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(F=1.\frac{32}{99}\)
\(F=\frac{32}{99}\)
Tính
F = \(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}.\frac{6^2}{5.7}.\frac{7^2}{6.8}.\frac{8^2}{7.9}\)
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Nhớ giải ra hết nhé
xin lỗi mình mới học lớp 5 thôi
Bài 1:Tính tổng
S=\(\frac{14}{3.5}+\frac{14}{5.7}+\frac{14}{7.9}+...+\frac{14}{61.63}\) so sánh với \(\frac{5}{2}\)
Help meeeeeeeeeeeeeeeee
Lớp 6
\(S=7(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{61}-\frac{1}{63}) \)
\(S=7(\frac{1}{3}-\frac{1}{63})\)
\(S=7(\frac{21}{63}-\frac{1}{63}) \)
\(S=7.\frac{20}{63}\)
\(S=\frac{20}{9}\)
Do đó:\(S<\frac{5}{2}\)
S=\(\frac{2.7}{3.5}+\frac{2.7}{5.7}+\frac{2.7}{7.9}+....+\frac{2.7}{61.63}\)và\(\frac{5}{2}\)
S=7.(\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....-\frac{1}{63}\)) và\(\frac{5}{2}\)
S=7.(\(\frac{1}{3}-\frac{1}{63}\)) và\(\frac{5}{2}\)
S=7.\(\frac{20}{63}\)và\(\frac{5}{2}\)
=>S=\(\frac{20}{9}\)so với \(\frac{5}{2}\)
=>S=\(\frac{40}{18}\)và\(\frac{45}{18}\)
=>S<\(\frac{5}{2}\)
a) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}+\frac{1}{90}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.10}\)
Ai làm hộ mình với ạ
a) 1/2 +1/6 +1/12 +...+1/72 +1/90
= 1/1.2 + 1/2.3 + 1/3.4 +....+1/8.9 + 1/9.10
= 1/1 -1/2 + 1/2 -1/3 + 1/3 -1/4 +....+1/9- 1/9 + 1/9 -1/10
= 1/1 -1/10
= 9/10
b) 2/3.5 + 2/5.7 + 2/7.9 + 2/9.10
= 2( 1/3.5 + 1/5.7 +1/7.9 + 1/9.10)
= 1/3-1/5 + 1/5- 1/7 +1/7-1/9 +1/9-1/10
= 1/3- /10
= 7/30