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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
\(A=\frac{1}{1.4.7}+\frac{1}{4.7.10}+...+\frac{1}{54.57.60}\)
\(\Rightarrow6A=\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\)
\(=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.47}-\frac{1}{57.60}\)
\(=\frac{1}{4}-\frac{1}{3420}=\frac{855}{3420}-\frac{1}{3420}=\frac{427}{1710}\)
\(\Rightarrow A=\frac{427}{1710}:6=\frac{427}{1710}.\frac{1}{6}=\frac{427}{10260}\)
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
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 = 18:26+(-5):27+(-22):86+12:39+(-32):43 = 9:13+(-5):27+(-11):43+4:13+(-32):43 = (9:13+4:13)+[(-11):43+(-32):43]+(-5):27 = 1+(-1)+5:27 = -5:27
B =(-10):12+8:15+(-19):56+3:(-18)+28:60 = (-5):6+8:15+(-19):56+1:(-6)+7:15 = [(-5):6+1:(-6)]+(8:15+7:15)+(-19):56 = (-1)+1+(-19):56 = (-19) :56
Bài 1:\(A=1-\frac{1}{2}+1-\frac{1}{6}+.......+1-\frac{1}{9900}\)
\(=1-\frac{1}{1.2}+1-\frac{1}{2.3}+........+1-\frac{1}{99.100}\)
\(=99-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\right)=99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=99-\left(1-\frac{1}{100}\right)=99-\frac{99}{100}=\frac{9801}{100}\)
Bài 2:\(A=\frac{1}{299}.\left(\frac{299}{1.300}+\frac{299}{2.301}+.........+\frac{299}{101.400}\right)\)
\(=\frac{1}{299}.\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+.........+\frac{1}{101}-\frac{1}{400}\right)\)
\(=\frac{1}{299}.\left(1+\frac{1}{2}+......+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-.......-\frac{1}{400}\right)\)
\(=\frac{1}{299}.\left[\left(1+\frac{1}{2}+.......+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+......+\frac{1}{400}\right)\right]\)(đpcm)
1/
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{9900}\right)\)
\(=\left(1+1+...+1\right)\left(50so\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\right)\)
\(=50-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=50-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=50-\left(1-\frac{1}{100}\right)=49+\frac{1}{100}=\frac{4901}{100}\)
2/
\(=\frac{1}{299}\left(\frac{299}{1.300}+\frac{299}{2.301}+...+\frac{299}{101.400}\right)\)
\(=\frac{1}{299}\left(1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\right)\)
\(=\frac{1}{299}\left[\left(1+\frac{1}{2}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400}\right)\right]\)
\(P=\frac{12}{1.4.7}+\frac{12}{4.7.10}+\frac{12}{7.10.13}+...+\frac{12}{54.57.60}\)
\(P=4.\left(\frac{3}{1.4.7}+\frac{3}{4.7.10}+\frac{3}{7.10.13}+...+\frac{3}{54.57.60}\right)\)
\(P=4\left(\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}\right)\)
\(P=4.\left(\frac{1}{4}-\frac{1}{3420}\right)\)
\(P=4.\frac{427}{1710}\)
\(P=\frac{854}{855}\)