1,B=\(\frac{3}{1.6}+\frac{3}{6.11}+\frac{3}{11.16}+..............+\frac{3}{96.101}\)
Những câu hỏi liên quan
b=\(\frac{3}{1.6}+\frac{3}{6.11}+\frac{3}{11.16}+..............+\frac{3}{96.101}\)
\(\Leftrightarrow B=\frac{3}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(\Leftrightarrow B=\frac{3}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(\Leftrightarrow B=\frac{3}{5}.\frac{100}{101}\)
\(\Leftrightarrow B=\frac{60}{101}\)
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Tính tổng\(S=\frac{3}{1.6}+\frac{3}{6.11}+\frac{3}{11.16}+...+\frac{3}{96.101}\)
\(.S=3.\left(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{96.101}\right)\)
\(\Rightarrow S=3.\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(\Rightarrow S=\frac{3}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(\Rightarrow S=\frac{3}{5}.\left(\frac{100}{101}\right)\)
\(S=\frac{60}{101}\)
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\(\frac{100}{101}\)nha
bạn tự tính
tíc mình nha
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S=3/1.6+3/6.11+3/11.16+...+3/96.101
=>S=1/1.6+1/6.11+1/11.16+...+1/96.101
S=1-1/6+1/6-1/11+1/11-1/16+...+1/96-1/101
S=1-1/101
S=100/101
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Tính tổng: S
\(\frac{3}{1.6}+\frac{3}{6.11}+\frac{3}{11.16}+..+\frac{3}{96.101}\)
\(\frac{3}{1.6}+\frac{3}{6.11}+\frac{3}{11.16}+...+\frac{3}{96.101}\)
\(=3.\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)\)
\(=\frac{3}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(=\frac{3}{5}.\left(1-\frac{1}{101}\right)\)
\(=\frac{3}{5}.\frac{100}{101}\)
\(=\frac{60}{101}\)
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tinh \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{96.101}\)
\(A=\frac{1}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{5}{96\cdot101}\right)\)
\(A=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(A=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(A=\frac{1}{5}\cdot\frac{100}{101}\)
\(A=\frac{20}{101}\)
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A = 1/5(1-1/6+1/6-1/11+1/11-1/16+.....+1/96-1/101)
= 1/5(1-1/101)=20/101
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tinh \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{96.101}\)
ta có : 1/1.6+1/6.11+1/11.16+....+1/96.101
= 1/5.5/1.6+ 1/5.5/6.11+1/5.5/11.16+...+1/5.5/96.101
=1/5 . ( 5/1.6+5/6.11+5/11.16+...+5/96.101)
=1/5 . ( 1/1-1/6 +1/6-1/11+1/11-1/16+....+1/96-1/101)
=1/5 . (1/1-1/101)
=1/5 . 100/101
= 20/101
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5A=\( 1-{1\over 6}+{1\over 6}-{1\over 11}+...{1\over 96}-{1\over 101}\)
=\(1- {1 \over 101}={100 \over 101}\)
suy ra A =\({20 \over 101}\)
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\(A=\frac{1}{5}\cdot\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+\frac{...5}{96\cdot101}\right)\)
\(A=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(A=\frac{1}{5}\left(\frac{1}{1}+0+0+0+...+0-\frac{1}{101}\right)\)
\(A=\frac{1}{5}\left(1-\frac{1}{101}\right)\)
\(A=\frac{1}{5}\cdot\frac{100}{101}\)
\(A=\frac{20}{101}\)
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C = \(\frac{1}{1.6}\)+ \(\frac{1}{6.11}\)+ \(\frac{1}{11.16}\)+ ........+ \(\frac{1}{96.101}\)
Giải dùm mik chút nhoa hôn hôn
\(C=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(C=\frac{1}{5}\left(1-\frac{1}{101}\right)\)
\(C=\frac{1}{5}.\frac{100}{101}=\frac{20}{101}\)
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\(5C=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\)
\(5C=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\)
\(5C=1-\frac{1}{101}\)
\(C=\frac{100}{\frac{101}{5}}\)
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Xem thêm câu trả lời
tính giá trị các biểu thức sau bằng cách hợp lýa, (6^3+ 3 x 6^2+ 3^3) : 13b, frac{8frac{3}{19}cdot5frac{1}{4}+3frac{16}{19}cdot5frac{1}{4}}{left(2frac{14}{17}-2frac{1}{34}right).34}c, frac{3}{1.6}+ frac{3}{6.11}+frac{3}{11.16}+....+frac{3}{96.101}làm câu nào cg đc cần gấp
Đọc tiếp
tính giá trị các biểu thức sau bằng cách hợp lý
a, (\(6^3\)+ 3 x \(6^2\)+ \(3^3\)) : 13
b, \(\frac{8\frac{3}{19}\cdot5\frac{1}{4}+3\frac{16}{19}\cdot5\frac{1}{4}}{\left(2\frac{14}{17}-2\frac{1}{34}\right).34}\)
c, \(\frac{3}{1.6}\)+ \(\frac{3}{6.11}\)+\(\frac{3}{11.16}\)+....+\(\frac{3}{96.101}\)
làm câu nào cg đc cần gấp
a)\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{2008.2011}\)\(\) b)\(\frac{-5}{1.6}-\frac{5}{6.11}-\frac{5}{11.16}-......-\frac{5}{2006.2011}\)
a,1/1-1/4+1/4-1/7+...+1/2008-1/2011
=(1-1/2011)+(-1/4+1/4)+...+(-1/2008+1/2008)
=1-1/2011+0+...+0
=1-1/2011
=2010/2011
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Tính nhanh:
a) \(P=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)
b) \(Q=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\)
\(b\)) \(Q=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)=\frac{150}{31}\)
\(a\)) Mình giải theo cách khác:
Chú ý rằng : \(\frac{3}{2.5}=\frac{1}{2}-\frac{1}{5};\frac{3}{5.8}=\frac{1}{5}-\frac{1}{8};\frac{3}{8.11}=\frac{1}{8}-\frac{1}{11};...;\frac{3}{17.20}=\frac{1}{17}-\frac{1}{20}\)
Do đó: \(P=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)