Bai 5. Tinh nhanh
a, \(\dfrac{1}{5.8}+\dfrac{1}{8.7}+\dfrac{1}{11.14}+.......+\dfrac{1}{605.606}\)
b,\(\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{11}-1\right)\left(\dfrac{1}{12}-1\right)....\left(\dfrac{1}{2012}-1\right)\)
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Tìm x:
a) \(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)
b) \(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(x+1\right):2}=1\dfrac{1991}{1993}\)
a)
<=> (1/3)[3/(5.8) + 3/(8.11) + ... + 3/[x(x+3)] = 101/1540
<=> (1/3)[(1/5 - 1/8) + (1/8 - 1/11) + ... + 1/x - 1/(x+3)] = 101/1540
<=> (1/3)[1/5 - 1/(x+3)] = 101/1540
<=> 1/5 - 1/(x+3) = 303/1540
<=> 1/(x+3) = 1/5 - 303/1540 = 5/1540 = 1/308
<=> x = 305
b)
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a)\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)
\(\dfrac{1.3}{5.8}+\dfrac{1.3}{8.11}+\dfrac{1.3}{11.14}+...+\dfrac{1.3}{x.\left(x+3\right)}=\dfrac{101.3}{1540}\)
\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}=\dfrac{303}{1540}\)
\(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}=\dfrac{1}{308}\)
308.1 = (x + 3).1
308 = x + 3
x = 308 - 3
x = 305
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a) \(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
=> \(3.\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{x\left(x+3\right)}\right)=\dfrac{101}{1540}.3\)
=> \(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{303}{1540}\)
=> \(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
=> \(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
=> \(\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}\)
=> \(\dfrac{1}{x+3}=\dfrac{1}{308}\)
=> \(x+3=308\)
=> \(x=305\)
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\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{y\left(y+3\right)}=\dfrac{98}{1545}\)
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{y\left(y+3\right)}=\dfrac{98}{1545}\)
\(\Leftrightarrow\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{y\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{y}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{y+3}{5\left(y+3\right)}-\dfrac{5}{5\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{y+3-5}{5\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{y-2}{5\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow515\left(y-2\right)=98.5\left(y+3\right)\)
\(\Leftrightarrow515y-1030=490y+1470\)
\(\Leftrightarrow25y-2500=0\\ \Leftrightarrow25y=2500\\ \Leftrightarrow y=100\)
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\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{y+3}\right)=\dfrac{98}{1545}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{1}{y+3}=\dfrac{1}{103}\)
hay x=100
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\(Tìmxbiết\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x+\left(x+3\right)}\)
Ta có: \(\dfrac{k}{x.\left(x+k\right)}=\dfrac{x+k-x}{x.\left(x+k\right)}=\dfrac{1}{x}-\dfrac{1}{x+k}\)
nên áp dụng ta có:
\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\)
\(=\dfrac{1}{5}-\dfrac{1}{x+3}\)
Nên $\dfrac{1}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\right)=\dfrac{1}{3}.(\dfrac{1}{5}-\dfrac{1}{x+3})$
Đến đây là làm được rồi nha
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Tính nhanh :
a) A = \(\dfrac{10\dfrac{1}{3}\left(26\dfrac{1}{3}-\dfrac{176}{7}\right)-\dfrac{12}{11}\left(\dfrac{10}{3}-1.75\right)}{\dfrac{5}{\left(91-0.25\right).\dfrac{60}{11}-1}}\)
b) ( 18.123+9.436.2+3.5310.6):(1+4+7+......+100-410)
Tính giá trị biểu thức
a, dfrac{5.8-5.6}{10} b, dfrac{left(-4right)^2}{5} B dfrac{3}{7}+left(dfrac{-1}{5}+dfrac{-3}{7}right) C left(6-2dfrac{4}{5}right).3dfrac{1}{8}-1dfrac{3}{5}:dfrac{1}{4}
D left(dfrac{-5}{24}+0,75+dfrac{7}{12}right):left(-2dfrac{1}{8}right) E dfrac{-5}{7}.dfrac{2}{11}+dfrac{-5}{7}.dfrac{9}{11}+1dfrac{5}{7} F dfrac{6}{7}+dfrac{5}{8}:5-dfrac{3}{16}.left(-2right)^2
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Tính giá trị biểu thức
a, \(\dfrac{5.8-5.6}{10}\) b, \(\dfrac{\left(-4\right)^2}{5}\) B = \(\dfrac{3}{7}+\left(\dfrac{-1}{5}+\dfrac{-3}{7}\right)\) C = \(\left(6-2\dfrac{4}{5}\right).3\dfrac{1}{8}-1\dfrac{3}{5}:\dfrac{1}{4}\)
D = \(\left(\dfrac{-5}{24}+0,75+\dfrac{7}{12}\right):\left(-2\dfrac{1}{8}\right)\) E = \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\) F = \(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2\)
a: \(=\dfrac{5\cdot\left(8-6\right)}{10}=\dfrac{5\cdot2}{10}=1\)
b: \(\dfrac{\left(-4\right)^2}{5}=\dfrac{16}{5}\)
\(B=\dfrac{3}{7}-\dfrac{1}{5}-\dfrac{3}{7}=-\dfrac{1}{5}\)
c: \(C=\left(6-2.8\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)
\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}\)
\(=5\cdot2-\dfrac{32}{5}=10-\dfrac{32}{5}=\dfrac{18}{5}\)
d: \(D=\left(\dfrac{-5}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{27}{24}\cdot\dfrac{-8}{17}=\dfrac{-9}{8}\cdot\dfrac{8}{17}=\dfrac{-9}{17}\)
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Bài 1 : Tính nhanh
A=\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
Bài 2:Tìm x biết
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\left(x+1\right):2}=1\dfrac{2007}{2009}\)
a/ \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+.......+\dfrac{1}{2^{10}}\)
\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.......+\dfrac{1}{2^9}\)
\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+......+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{10}}\right)\)
\(\Leftrightarrow A=1-\dfrac{1}{2^{10}}\)
b/ \(\dfrac{1}{5.8}+\dfrac{1}{8.11}+.......+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(\Leftrightarrow3\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+......+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{101}{1540}.3\)
\(\Leftrightarrow\dfrac{3}{5.8}+\dfrac{3}{8.11}+......+\dfrac{3}{x\left(x+3\right)}=\dfrac{303}{1540}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+.....+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)
\(\Leftrightarrow x+3=308\)
\(\Leftrightarrow x=305\)
Vậy ..
c/ \(1+\dfrac{1}{3}+\dfrac{1}{6}+........+\dfrac{1}{x\left(x+1\right):2}=1\dfrac{2007}{2009}\)
\(\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{6}+.......+\dfrac{1}{x\left(x+1\right):2}\right)=\dfrac{4016}{2009}.\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+......+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)
\(\Leftrightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+......+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2009}\)
\(\Leftrightarrow x+1=2009\)
\(\Leftrightarrow x=2008\)
Vậy ..
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bài 1:
A=\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
ta thấy 2A=\(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^9}\)
=>2A-A=\(1-\dfrac{1}{2^{10}}=\dfrac{1023}{1024}\)
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2.Tính:
Aleft(dfrac{1}{10}-1right).left(dfrac{1}{11}-1right)left(dfrac{1}{12}-1right).....left(dfrac{1}{100}-1right)
Bleft(1-dfrac{1}{2^2}right)left(1-dfrac{1}{3^2}right)left(1-dfrac{1}{10^2}right)
Cleft(dfrac{7}{9}+1right)left(dfrac{7}{20}+1right)left(dfrac{7}{33}+1right)...left(dfrac{7}{108080}+1right)
Dleft(1-dfrac{28}{10}right)left(1-dfrac{52}{22}right)left(1-dfrac{80}{36}right)...left(1-dfrac{21808}{10900}right)
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2.Tính:
A=\(\left(\dfrac{1}{10}-1\right).\left(\dfrac{1}{11}-1\right)\left(\dfrac{1}{12}-1\right).....\left(\dfrac{1}{100}-1\right)\)
B=\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{10^2}\right)\)
C=\(\left(\dfrac{7}{9}+1\right)\left(\dfrac{7}{20}+1\right)\left(\dfrac{7}{33}+1\right)...\left(\dfrac{7}{108080}+1\right)\)
D=\(\left(1-\dfrac{28}{10}\right)\left(1-\dfrac{52}{22}\right)\left(1-\dfrac{80}{36}\right)...\left(1-\dfrac{21808}{10900}\right)\)
Tính hợp lý:
\(A=\dfrac{7}{12}+\dfrac{5}{12}:6-\dfrac{11}{36}\) \(B=\left(\dfrac{4}{5}+\dfrac{1}{2}\right):\left(\dfrac{3}{13}-\dfrac{8}{13}\right)\)
\(C=\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right):\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right)\)
a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)
c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
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Tính:
a) \(\dfrac{5}{16}-\dfrac{5}{24};\)
b) \(\dfrac{2}{11}+\left(\dfrac{-5}{11}-\dfrac{9}{11}\right)\);
c) \(\dfrac{1}{10}-\left(\dfrac{5}{12}-\dfrac{1}{15}\right).\)
\(a.\)
\(\dfrac{5}{16}-\dfrac{5}{24}=\dfrac{5\cdot3-5\cdot2}{48}=\dfrac{15-10}{48}=\dfrac{5}{48}\)
\(b.\)
\(\dfrac{2}{11}+\left(\dfrac{-5}{11}-\dfrac{9}{11}\right)=\dfrac{2-5-9}{11}=-\dfrac{12}{11}\)
\(c.\)
\(\dfrac{1}{10}-\left(\dfrac{5}{12}-\dfrac{1}{15}\right)=\dfrac{1}{10}-\dfrac{5}{12}+\dfrac{1}{15}=\dfrac{6-5\cdot5+4}{60}=-\dfrac{15}{60}=-\dfrac{1}{4}\)
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a) \(\dfrac{5}{16}-\dfrac{5}{24}=\dfrac{15}{48}-\dfrac{10}{48}=\dfrac{15-10}{48}=\dfrac{5}{48}\)
b)\(\dfrac{2}{11}+\left(\dfrac{-5}{11}-\dfrac{9}{11}\right)=\dfrac{2}{11}-\dfrac{5}{11}-\dfrac{9}{11}=\dfrac{2-5-9}{11}=\dfrac{-12}{11}\)
c)\(\dfrac{1}{10}-\left(\dfrac{5}{12}-\dfrac{1}{15}\right)=\dfrac{1}{10}-\dfrac{7}{20}=\dfrac{2}{20}-\dfrac{7}{20}=\dfrac{-5}{20}=\dfrac{-1}{4}\)
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