A=1/2x9+1/4x15+1/6x21+...+1/98x297
Chứng minh rằng : 1/15<A<2/15
A=1/2x9 + 1/4x15 + 1/6x21 + ... +1/98x297
Chứng minh rằng 1/15 <A< 2/15
1/3x12 + 1/4x15 + 1/5x18 +......+ 1/59x180
Em nên gõ công thức trực quan để đề bài rõ ràng nhé
Tính a)16/13-(3/15-6/13)
b)21/8-(1/2+3/5)
c)3/2x9/5-3-2x2/7
d)32/45x18/16
e)(4/3-4/5)x25/2
f)2-1/3-1/2-1/6-3/12
a)\(=\dfrac{16}{13}-\dfrac{3}{15}+\dfrac{6}{13}=\dfrac{22}{13}-\dfrac{3}{15}=\dfrac{96}{65}\)
b)\(=\dfrac{21}{8}-\left(\dfrac{5}{10}+\dfrac{6}{10}\right)=\dfrac{21}{8}-\dfrac{11}{10}=\dfrac{61}{40}\)
c)\(=\dfrac{27}{10}-3-\dfrac{4}{7}--\dfrac{61}{70}\)
d)\(=\dfrac{576}{702}=\dfrac{4}{5}\)
e)\(=\left(\dfrac{20}{15}-\dfrac{12}{15}\right)\times\dfrac{25}{2}=\dfrac{8}{15}\times\dfrac{25}{2}=\dfrac{20}{3}\)
f)\(=\dfrac{24}{12}-\dfrac{4}{12}-\dfrac{6}{12}-\dfrac{2}{12}-\dfrac{3}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
d) \(=\dfrac{32}{45}x\dfrac{18}{16}=\dfrac{576}{720}=\dfrac{115}{114}\)
e) \(=\left(\dfrac{4}{3}-\dfrac{4}{5}\right)x\dfrac{25}{12}=\dfrac{8}{15}x\dfrac{25}{12}=\dfrac{200}{180}=\dfrac{10}{9}\)
f) \(=\dfrac{2}{1}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{3}{12}=\dfrac{24}{12}-\dfrac{4}{12}-\dfrac{6}{12}-\dfrac{2}{12}-\dfrac{3}{12}\)
\(=\dfrac{20}{12}-\dfrac{4}{12}-\dfrac{3}{12}=\dfrac{16}{12}-\dfrac{3}{12}=\dfrac{13}{12}\)
X : 1/2 = 2/3x4/5x3/4x15/8
x:1/2 = (2/3.3/4).4/5.15/8
x:1/2 = 6/12.60/40
x:1/2 = 1/2.3/2
x:1/2 = 3/4
x = 3/4.1/2
x = 3/8
(. là nhân nhé mik đánh máy sách tay nên ko có dấu nhân k mk nhé)
1+3++6+10+15+..........+45+55
1x10+2x9+3x8+...........+8x3+9x2+10x1
1,Chứng minh rằng: 1<1/5+1/6+1/7+....+1/17<2
2,Cho A=1/2× 3/4×5/6×....×99/100
Chứng minh rằng 1/15<A<1/10
\(A=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{9}+\dfrac{1}{10}}{\dfrac{1}{1x10}+\dfrac{1}{2x9}+\dfrac{1}{3x8}+...+\dfrac{1}{8x3}+\dfrac{1}{9x2}+\dfrac{1}{10x1}}\)
Cho A=1/10+1/15+1/21+...+1/120
hãy chứng minh rằng A<1
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\)
Ta có :
\(\frac{1}{10}< 1\)
\(\frac{1}{15}< 1\)
\(\frac{1}{21}< 1\)
........................
\(\frac{1}{120}< 1\)
\(\Rightarrow\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}< 1\)
\(\Rightarrow A< 1\)( đpcm)
Ta có : A = \(\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)
= \(\frac{1}{20}\times2+\frac{1}{30}\times2+...+\frac{1}{240}\times2\)
= \(2\times\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)
= \(2\times\left(\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{15\times16}\right)\)
= \(2\times\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)
= \(2\times\left(\frac{1}{4}-\frac{1}{16}\right)\)
= \(2\times\frac{3}{16}\)
= \(\frac{3}{8}\)< 1
=> A < 1
a/ 1/2x9 + 1/9x7+1/7x19+...........+1/252x509
b/1/10x9+1/18x13+1/26x17+..................+1/802x405
Ai BIẾT CÂU NÀY GIÚP MIK VỚI
a) \(\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{202.509}=\frac{2}{4.9}+\frac{2}{9.14}+\frac{2}{14.19}+...+\frac{2}{504.509}\)
\(=\frac{2}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{504.509}\right)\)
\(=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{504}-\frac{1}{509}\right)=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{509}\right)\)
\(=\frac{2}{5}.\frac{505}{2036}=\frac{101}{1018}\)
b) \(\frac{1}{10.9}+\frac{1}{18.13}+...+\frac{1}{802.405}=\frac{2}{10.18}+\frac{2}{18.26}+...+\frac{2}{802.810}\)
\(=\frac{2}{8}\left(\frac{8}{10.18}+\frac{8}{18.26}+...+\frac{8}{802.810}\right)=\frac{1}{4}\left(\frac{1}{10}-\frac{1}{18}+\frac{1}{18}-\frac{1}{26}+...+\frac{1}{802}-\frac{1}{810}\right)\)
\(=\frac{1}{4}\left(\frac{1}{10}-\frac{1}{810}\right)=\frac{1}{4}.\frac{40}{405}=\frac{10}{405}\)
Bạn vào câu hỏi tương tự tham khảo !