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\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right)+\dfrac{4}{5}\\ =-\dfrac{5}{21}:\dfrac{4}{5}+\dfrac{5}{21}\\ =\left(-\dfrac{5}{21}+\dfrac{5}{21}\right):\dfrac{4}{5}\\ =0:\dfrac{4}{5}\\ =0.\)
Sửa cho mk dòng đầu là :4/5 và dòng tiếp theo mk thiếu :4/5
\(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+..+\dfrac{1}{44.49}\right)\left(\dfrac{1-3-5-7-..-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+..+\dfrac{5}{44.49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\)
\(A=\dfrac{9}{196}\left(\dfrac{1-3-5-7-...-49}{89}\right)\)
Ta đặt: \(P=1-3-5-7-...-49\\ =1-\left(3+5+7+..+49\right)\\ =1-624\\ =-623\\ \Rightarrow\dfrac{9}{196}.-\dfrac{623}{89}=-\dfrac{9}{28}.\)
Ta có: �=(14⋅9+19⋅14+114⋅19+...+144⋅49)⋅1−3−5−7−...−4989A=(4⋅91+9⋅141+14⋅191+...+44⋅491)⋅891−3−5−7−...−49
⇔�=15⋅(54⋅9+59⋅14+514⋅19+...+544⋅49)⋅1−3−5−7−...−4989⇔A=51⋅(4⋅95+9⋅145+14⋅195+...+44⋅495)⋅891−3−5−7−...−49
⇔�=15⋅(14−19+19−114+114−119+...+144−149)⋅1−3−5−7−...−4989⇔A=51⋅(41−91+91−141+141−191+...+441−491)⋅891−3−5−7−...−49
⇔�=15⋅(14−149)⋅1−3−5−7−...−4989⇔A=51⋅(41−491)⋅891−3−5−7−...−49
⇔�=15⋅(49−44⋅49)⋅1−3−5−7−...−4989⇔A=51⋅(4⋅4949−4)⋅891−3−5−7−...−49
⇔�=15⋅45196⋅1−3−5−7−...−4989⇔A=51⋅19645⋅891−3−5−7−...−49
⇔�=9196⋅1−3−5−7−...−4989⇔A=1969⋅891−3−5−7−...−49
⇔�=9196⋅−62389=−928⇔A=1969⋅89−623=−289
\(A=\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right)....\left(\dfrac{1}{100}-1\right).\)
\(\Rightarrow A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)\)
mà A có 9 dấu - \(\left(4;9;16;25;36;49;64;81;100\right)\)
\(\Rightarrow0>A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)=-\dfrac{1}{2}\)
Ta lại có \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{21}{42}\\\dfrac{11}{21}=\dfrac{22}{42}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2}< \dfrac{11}{21}\Rightarrow-\dfrac{1}{2}>-\dfrac{11}{21}\)
\(\Rightarrow A>-\dfrac{11}{21}\)
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)\)
\(A=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)...\left(-\dfrac{10^2-1}{10^2}\right)\)
\(A=\left[-\dfrac{1\cdot3}{2\cdot2}\right]\left[-\dfrac{2\cdot4}{3\cdot3}\right]...\left[-\dfrac{9\cdot11}{10\cdot10}\right]\)
Dễ thấy A có 9 thừa số, suy ra
\(A=-\dfrac{1\cdot3\cdot2\cdot4\cdot...\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot...\cdot10.10}=-\dfrac{1\cdot11}{2\cdot10}=\dfrac{-11}{20}\)
Vì 20 < 21 nên \(\dfrac{11}{20}>\dfrac{11}{21}\), suy ra \(\dfrac{-11}{20}< \dfrac{-11}{21}\)
Vậy \(A< \dfrac{-11}{21}\)
a) \(\left(1\dfrac{1}{2}\right)\left(1\dfrac{1}{3}\right)..............\left(1\dfrac{1}{100}\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}....................\dfrac{101}{100}\)
\(=\dfrac{1}{2}.\dfrac{101}{1}=\dfrac{101}{2}\)
b) \(1\dfrac{1}{2}.1\dfrac{1}{3}.1\dfrac{1}{4}...................1\dfrac{1}{2007}\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}....................\dfrac{2008}{2007}\)
\(=\dfrac{1}{2}.\dfrac{2008}{1}=1004\)
c) \(1\dfrac{1}{2}.1\dfrac{1}{3}.....................1\dfrac{1}{2017}\)
\(=\dfrac{3}{2}.\dfrac{4}{3}..................\dfrac{2018}{2017}\)
\(=\dfrac{1}{2}.\dfrac{2018}{1}=1009\)
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{125}{376}\)
\(\Leftrightarrow\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{125}{376}\)
\(\Leftrightarrow\dfrac{1}{3}\left(1-\dfrac{1}{x+3}\right)=\dfrac{125}{376}\left(x\ne0;x\ne-3\right)\)
\(\Leftrightarrow\dfrac{x+3-1}{x+3}=\dfrac{3.125}{376}\Leftrightarrow\dfrac{x+2}{x+3}=\dfrac{3.125.}{376}.\dfrac{\left(x+3\right)}{x+3}\)
\(\Leftrightarrow376\left(x+2\right)=3.125.\left(x+3\right)\)
\(\Leftrightarrow376x+752=375x+1125\)
\(\Leftrightarrow376x-375x=1125-752\Leftrightarrow x=373\left(x\in N^{\cdot}\right)\)
\(\dfrac{6^3+2.6^2+2^3}{37}=\dfrac{2^3.3^3+2.2^2.3^2+2^3}{37}\\ \\ \\ \\ \\ \\ \\ \\ \\=\dfrac{2^3.3^3+2^3.3^2+2^3}{37}\\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{2^3.\left(3^3+3^2+1\right)}{37}=\dfrac{2^3.37}{37}=2^3=8\)
\(\dfrac{6^3+2.6^2+2^3}{37}=\dfrac{216+72+8}{37}=\dfrac{296}{37}=8\)
\(a)\dfrac{-5}{21}-\dfrac{1}{3}+3\dfrac{1}{2}.\left(\dfrac{-2}{3}\right)^3\)
\(=\dfrac{-5}{21}+\dfrac{-7}{21}+\dfrac{7}{2}.\dfrac{-8}{27}\)
\(=-\dfrac{4}{7}+\dfrac{-28}{27}\)
\(=\dfrac{-108}{189}+\dfrac{-196}{189}\)
\(=-\dfrac{304}{189}\)
\(b)-2\dfrac{1}{3}+\left(\dfrac{3}{8}-\dfrac{3}{4}\right)^3:\dfrac{5}{9}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\left(\dfrac{3}{8}-\dfrac{6}{8}\right)^3.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\left(-\dfrac{3}{8}\right)^3.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\dfrac{-27}{512}.\dfrac{9}{5}-\dfrac{1}{2}\)
\(=-\dfrac{7}{3}+\dfrac{-243}{2560}-\dfrac{1}{2}\)
\(=\dfrac{-17920}{7680}+\dfrac{-729}{7680}+\dfrac{-3840}{7680}\)
\(=\dfrac{-22489}{7680}\)
1, \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...........\left(1-\dfrac{1}{n+1}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right)...........\left(\dfrac{n+1}{n+1}-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}..............\dfrac{n}{n+1}\)
\(=\dfrac{1.2.3........n}{2.3.......\left(n+1\right)}\)
\(=\dfrac{1}{n+1}\)
2, \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...........+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+............+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
C=\(-66\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)
=\(-66.\left(\dfrac{5}{66}\right)+124\left(-37-63\right)=-5+124.\left(-100\right)\)
=-12405
(1/2-1)(1/3-1)(1/4-1)...(1/2003-1)
=(-1)/2.(-2)/3.(-3)/4....(-2002)/2003
=\(\frac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-2002\right)}{2.3.4....2003}=-\frac{1}{2003}\)
\(\left(\dfrac{-1}{3}\right)^{-1}=-3\)
Vì \(\left(\dfrac{-1}{3}\right)^{-1}=\left(\dfrac{1}{-3}\right)^{-1}=\left(-3^{-1}\right)^{-1}=-3^{-1\times\left(-1\right)}=-3^1=-3\)
=> \(\left(\dfrac{-1}{3}\right)^{-1}=-3\)