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Ta quy đồng tử để có cùng tử là 3 :
\(\frac{1}{7}=\frac{3}{21}\)
\(\frac{1}{8}=\frac{3}{24}\)
=>\(\frac{3}{21}< x< \frac{3}{24}\)
Nên \(x=\frac{3}{22};\frac{3}{23}\)
Vậy tổng các phân số lớn hơn \(\frac{1}{7}\)và nhỏ hơn \(\frac{1}{8}\)là \(\frac{135}{506}\)
k mình nha các bạn và mình chúc các bạn học giỏi nha
1
B= 12/1.4.7 + 12/4.7.10 + 12/7.10.13 + ... + 12/54.57.60
=> 1/2B= 6/1.4.7 + 6/4.7.10 + 6/7.10.13 + ... + 6/54.57.60
=> 1/2B = 1/1.4 - 1/4.7 +1/4.7 - 1/7.10 +1/7.10 - 1/10.13 + ... + 1/54.57 - 1/57.60
=> 1/2B =1/1.4 - 1/57.60
=> 1/2B = 1/4 - 1/3420
=> 1/2B = 427/1710
=> B = 427/1710 . 2
=> B = 427/855
2
A= 1+ 1/22 + 1/32 +...+1/1002
=1+ 1/2.2 + 1/3.3 +...+ 1/100.100
=> A< 1+ 1/1.2 + 1/2.3 +...+ 1/99.100
= 1+ 1 - 1/2 +1/2 - 1/3 +...+1/99 - 1/100
= 2- 1/100 < 2
Vậy A < 2
Ta có \(A=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+\frac{1}{7.10}-...+\frac{1}{16.19}-\frac{1}{19.22}\)
\(=\frac{1}{4}-\frac{1}{418}=\frac{207}{836}\)
\(A=\frac{6}{1\cdot4\cdot7}+\frac{6}{4\cdot7\cdot10}+\frac{6}{7\cdot10\cdot13}+...+\frac{6}{16\cdot19\cdot22}\)
\(A=\frac{1}{1\cdot4}-\frac{1}{4\cdot7}+\frac{1}{4\cdot7}-\frac{1}{7\cdot10}+...+\frac{1}{16\cdot19}-\frac{1}{19\cdot22}\)
\(A=\frac{1}{4}-\frac{1}{19\cdot22}=\frac{207}{836}\)
\(A=\frac{1}{1.4.7}+\frac{1}{4.7.10}+...+\frac{1}{54.57.60}\)
\(\Rightarrow6A=\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\)
\(=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.47}-\frac{1}{57.60}\)
\(=\frac{1}{4}-\frac{1}{3420}=\frac{855}{3420}-\frac{1}{3420}=\frac{427}{1710}\)
\(\Rightarrow A=\frac{427}{1710}:6=\frac{427}{1710}.\frac{1}{6}=\frac{427}{10260}\)
Nhận thấy:
\(\frac{6}{1.4.7}=\frac{1}{1.4}-\frac{1}{4.7}\)
...............
\(\frac{6}{54.57.60}=\frac{1}{54.57}-\frac{1}{57.60}\)
=> ta phải nhân A vói 6
=> 6A =
\(\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}=\frac{1}{4}-\frac{1}{57.60}=\frac{427}{1710}\)
=> A = 427/1710 : 6 =427/10260
BT 7 :
\(P=\frac{12}{1.4.7}+\frac{12}{4.7.10}+\frac{12}{7.10.13}+...+\frac{12}{54.57.60}\)
\(P=\frac{12}{6}\left(\frac{6}{1.4.7}+\frac{6}{4.7.10}+\frac{6}{7.10.13}+...+\frac{6}{54.57.60}\right)\)
\(P=2\left(\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+\frac{1}{7.10}-\frac{1}{10.13}+...+\frac{1}{54.57}-\frac{1}{57.60}\right)\)
\(P=2\left(\frac{1}{1.4}-\frac{1}{57.60}\right)\)
\(P=2\left(\frac{1}{4}-\frac{1}{3420}\right)\)
\(P=\frac{1}{2}-\frac{1}{1710}< \frac{1}{2}\)
Vậy \(P< \frac{1}{2}\)
Chúc bạn học tốt ~
\(B=81.\left(\frac{12-\frac{12}{7}-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right).\frac{158158158}{711711711}\)
\(\Leftrightarrow B=81.\left(\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right).\frac{158\left(1001001\right)}{711\left(1001001\right)}\)
\(\Leftrightarrow B=81\left(\frac{12}{3}:\frac{5}{6}\right).\frac{158}{711}\)
\(\Leftrightarrow B=81\left(3.\frac{6}{5}\right).\frac{2}{9}\)
\(\Leftrightarrow B=81.\frac{18}{5}.\frac{2}{9}\)
\(\Leftrightarrow B=\frac{324}{5}\)
Hok tốt!!
\(P=\frac{12}{1.4.7}+\frac{12}{4.7.10}+\frac{12}{7.10.13}+...+\frac{12}{54.57.60}\)
\(P=4.\left(\frac{3}{1.4.7}+\frac{3}{4.7.10}+\frac{3}{7.10.13}+...+\frac{3}{54.57.60}\right)\)
\(P=4\left(\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}\right)\)
\(P=4.\left(\frac{1}{4}-\frac{1}{3420}\right)\)
\(P=4.\frac{427}{1710}\)
\(P=\frac{854}{855}\)