Tính \(D=\frac{2}{45}+\frac{1}{99}-\frac{2}{77}+\frac{4}{105}\)
Những câu hỏi liên quan
Tính: \(F=\frac{2}{15}+\frac{2}{35}+\frac{2}{99}+\frac{4}{77}\)
Tính D biết : \(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}}\)
Xét mẫu số:
\(A=\frac{100-1}{1}+\frac{100-2}{2}+\frac{100-3}{3}+.......+\frac{100-99}{99}\)
\(\Rightarrow A=\left(\frac{100}{1}+\frac{100}{2}+....+\frac{100}{99}\right)-\left(\frac{1}{1}+\frac{2}{2}+....+\frac{99}{99}\right)\)
\(\Rightarrow A=100+100.\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{99}\right)-99\)
\(A=1+100.\left(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)=100.\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\right)\)
Vậy \(D=\frac{1}{100}\)
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26 tháng 3 2016 lúc 16:34
Tính dãy số sau :
\(D=\frac{100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{98}{99}+\frac{99}{100}}\)
Cho \(P=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+..+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{100}}\)và \(Q=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-..-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+..+\frac{1}{500}}\)
a)Tính P,Q b) Tính tỉ số % của P và 3Q
Bài 1Tìm x biếtx-frac{37}{45}frac{4}{5.9}+frac{4}{9.13}+...+frac{4}{41.45}Bài 2 Tính giá trị các biểu thức sau a) A frac{1+frac{1}{3}+frac{1}{5}+...+frac{1}{97}+frac{1}{99}}{frac{1}{1.99}+frac{1}{3.97}+frac{1}{5.95}+...+frac{1}{97.3}+frac{1}{99.1}}b) B frac{frac{1}{2}+frac{1}{3}+frac{1}{4}+...+frac{1}{100}}{frac{99}{1}+frac{98}{2}+frac{97}{3}+...+frac{1}{99}}
Đọc tiếp
Bài 1Tìm x biết
\(x-\frac{37}{45}=\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\)
Bài 2 Tính giá trị các biểu thức sau
a) A = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{97.3}+\frac{1}{99.1}}\)
b) B = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}}\)
\(x-\frac{37}{45}=\frac{4}{5.9}+\frac{4}{9.13}+.....+\frac{4}{41.45}\)
\(\Rightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)
\(\Rightarrow x-\frac{37}{45}=\frac{1}{5}-\frac{1}{45}\)
\(\Rightarrow x-\frac{37}{45}=\frac{8}{45}\)
\(\Rightarrow x=\frac{37}{45}+\frac{8}{45}\)
\(\Rightarrow x=1\)
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Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)
\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)
\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
\(A=\frac{B}{6}=\frac{100}{2}=50\)
Vậy \(A=50\)
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Tính nhanh:
a)\(\frac{5}{30}+\frac{15}{90}+\frac{25}{150}+\frac{35}{210}+\frac{45}{270}\)
b)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}\)
c)\(\frac{1}{15}+\frac{4}{30}+\frac{2}{45}+\frac{16}{60}+\frac{25}{75}+\frac{36}{90}+\frac{49}{105}+\frac{64}{120}+\frac{81}{135}\)
a) 5/30+15/90+25/150+35/210+45/270
=1/6+1/6+1/6+1/6+1/6
=1/6 x 5
=5/6
b) 1/2+1/6+1/12+1/20+....+1/56
=1/1x2+1/2x3+1/3x4+1/4x5+.....1/7x8
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......-1/7+1/7-1/8
=1/1-1/8
=7/8
c) mình chịu
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chịu,khó quá,nhất là câu c ý
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Xem thêm câu trả lời
21 tháng 7 2018 lúc 10:28
Tính
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2004\times2005}+\frac{1}{2005\times2006}=A\)
\(\frac{1}{6}+\frac{2}{15}+\frac{4}{45}+\frac{2}{99}+\frac{10}{600}=A\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2004\cdot2005}+\frac{1}{2005\cdot2006}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2004}-\frac{1}{2005}+\frac{1}{2005}-\frac{1}{2006}\)
\(A=1-\frac{1}{2006}=\frac{2005}{2006}\)
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\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2005.2006}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(\Rightarrow A=1-\frac{1}{2006}\)
\(\Rightarrow A=\frac{2005}{2006}\)
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\(1)A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2004.2005}+\frac{1}{2005.2006}\)
\(\implies A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2004}-\frac{1}{2005}+\frac{1}{2005}-\frac{1}{2006}\)
\(\implies A=1-\frac{1}{2006}\)
\(\implies A=\frac{2005}{2006}\)
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Xem thêm câu trả lời
A=\(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+....+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}}\)
B=\(\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-....-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+....+\frac{1}{500}}\)
cho \(M=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\) và \(N=\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
Tính tỉ số M với N