\(\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+\frac{5}{42}+.....+\frac{5}{9900}\)
Tính :
\(\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+...+\frac{5}{9900}=?\)
A = 5 x (\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{9900}\))
A = 5 x ( \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{99}-\frac{1}{100}\))
A = 5x( \(\frac{1}{2}-\frac{1}{100}\))
A = \(\frac{49}{20}\)
Gọi tổng trên là A
\(\Leftrightarrow A=5\times\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)\)
(Tính dãy trong ngoặc) Gọi dãy trong ngoặc là B
\(\Leftrightarrow2B=\frac{1}{3}+\frac{1}{6}+...+\frac{1}{4950}\)
\(\Leftrightarrow2B-B=\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{4950}\right)-\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)\)
\(\Leftrightarrow B=\frac{1}{3}-\frac{1}{9900}+0+...+0\)
\(\Leftrightarrow B=\frac{3299}{9900}\)
\(\Rightarrow A=5\times\frac{3299}{9900}\)
\(\Rightarrow A=1,6661616...\approx1,7\)
=5/2 x 3 + 5/3x4 + 5/4 x5 + 5/5 x6 + ....... + 5/99 x 100
= 5 x ( 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 +...........+ 1/99 - 1/100)
= 5 x ( 1/2 - 1/100)
= 5 x 49/100
= 49/20
a) \(\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+...+\frac{5}{9900}\)
b) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
c) \(\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+...+\frac{91}{90}\)
d) \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{2}{3}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
Mấy câu như này tách ra kiểu gì?
\(\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+...+\frac{5}{9900}=\frac{5}{3.4}+\frac{5}{4.5}+\frac{5}{5.6}+...+\frac{5}{99.100}\)
\(5\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(5\left(\frac{1}{3}-\frac{1}{100}\right)=\frac{97}{60}\)
\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+...+\frac{71}{72}+\frac{89}{90}=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)=8-\frac{2}{5}=\frac{38}{5}\)
Tính giá trị của biểu thức:
\(A=\frac{5}{2}+\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+\frac{5}{42}+\frac{5}{56}+\frac{5}{72}\)
Đây nha! Tick cho mik nhé! Chỗ mik xuống dòng là viết tiếp đó!
Xin lỗi nhé hôm qua học thêm hơi vội!!! Sửa 8/9:5 thành 8/9.5=40/9
Tính tổng dãy số sau : \(A=\frac{5}{2}+\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+....+\frac{5}{9506}+\frac{5}{9702}+\frac{5}{9900}\)
A= 5.(1/2 + 1/6+1/12+1/20+...+1/9506+1/9702+1/9900)
= 5. (1/1.2 + 1/2.3+1/3.4+1/4.5+...1/97.98+1/98.99+1/99.100)
= 5 .(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/97-1/98+1/98-1/99+1/99-1/100)
= 5.(1-1/100)=5. 99/100=99/20
Tính nhanh:
I=\(\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+\frac{5}{42}+\frac{5}{56}+\frac{5}{72}+\frac{5}{90}\)
Ai nhanh mik tick cho mơn!
\(I=\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+...+\frac{5}{90}\)( viết tắt )
\(I=\frac{5}{2.3}+\frac{5}{3.4}+\frac{5}{4.5}+...+\frac{5}{9.10}\)
\(I=5\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(I=5\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(I=5\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(I=5\times\frac{2}{5}\)
\(I=2\)
Vậy \(I=2\)
Tk nha bn ~~
\(I=\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+\frac{5}{42}+\frac{5}{56}+\frac{5}{72}+\frac{5}{90}\)
\(I=\frac{5}{2\cdot3}+\frac{5}{3\cdot4}+\frac{5}{4\cdot5}+\frac{5}{5\cdot6}+\frac{5}{6\cdot7}+\frac{5}{7\cdot8}+\frac{5}{8\cdot9}+\frac{5}{9\cdot10}\)
\(I=5\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\)
Theo tính chất của toán HSG lớp 6, ta được
\(I=5\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(I=5\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(I=5\left(\frac{5}{10}-\frac{1}{10}\right)\)
\(I=5\cdot\frac{4}{10}=5\cdot\frac{2}{5}=\frac{10}{5}=2\)
\(I=\frac{5}{6}+\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+\frac{5}{42}+\frac{5}{56}+\frac{5}{72}+\frac{5}{90}\)
\(I=5\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(I=5\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(I=5\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(I=5\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(I=5.\frac{2}{5}\)
\(I=2\)
Cho A=\(\frac{3}{2}+\frac{13}{12}+\frac{21}{20}+...+\frac{9901}{9900}\)và B=\(\frac{5}{6}+\frac{19}{20}+\frac{41}{42}+...+\frac{10099}{10100}\)Tính A-B
Tính A = \(\frac{\frac{5}{2}-\frac{5}{7}+\frac{10}{5}}{\frac{1}{2}+\frac{1}{7}-\frac{1}{5}}-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
Ai dảnh kb nha
A=1+2+3+4+5+...+99+100
B=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9900}\)
A=1+2+3+4+5+...+99+100
A=(1+100).100:2=101.50=5050
B=1/2+1/6+1/12+1/20+1/30+...+1/9900
B=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+....+1/99.100
B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/99-1/100
B=1-1/100=99/100
A = 100 x 101 : 2 = 5050
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
S=\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{42}\)
\(S=\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{42}\)
\(=1+\frac{1}{2}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)+ \(\frac{1}{6}-\frac{1}{6}-\frac{1}{7}+\frac{1}{7}\)+ \(\frac{1}{8}-\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{9}{9}-\frac{1}{9}\)
\(=\frac{8}{9}\)
Chúc bạn học tốt !!!