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bài 1
a,\((\)\(\dfrac{-4}{21}\)\()\)x =\(\dfrac{28}{3}\)\(\times\)\(\dfrac{3}{28}\)
\(\Leftrightarrow\)\(\dfrac{-4}{21}\) x =1
\(\Rightarrow\)x = \(\dfrac{-21}{4}\)
b, \(\dfrac{17}{33}\)x = \(\dfrac{1}{56}\)\(\times\)56
\(\Leftrightarrow\)\(\dfrac{17}{33}\)x = 1
\(\Rightarrow\)x = \(\dfrac{33}{17}\)
bài 2 :
a, A=\(\dfrac{25}{32}\)
số nghịch đảo của A là \(\dfrac{32}{25}\)
B=\(\dfrac{3}{7}\)
số nghịch đảo của B là \(\dfrac{7}{3}\)
b, gọi tổng hai số nghịch đảo 2 số đó là Q
Q= \(\dfrac{32}{25}\) +\(\dfrac{7}{3}\)=\(\dfrac{271}{75}\)
a) 1619 và 825
Ta có :
1619 = ( 24 )19 = 276
825 = ( 23 )25 = 275
Vì 276 > 275 Nên 1619 > 825
b) 536 và 1124
Ta có :
536 = ( 53 )12 = 12512
1124 = ( 112 )12 = 12112
Vì 12512 > 12112 Nên 536 > 1124
1.
\(M=3^0+3^1+......+3^{50}.\)
\(\Rightarrow3M=3+3^2+.......+3^{51}\)
\(\Rightarrow3M-M=\left(3+3^2+.......+3^{51}\right)-\left(3^0+3+.....+3^{50}\right)\)
\(\Rightarrow2M=3^{51}-1\)
\(\Rightarrow M=\frac{3^{51}-1}{2}\)
2.
\(a,\)Ta có : \(16^{19}=\left(2^4\right)^{19}=2^{76}\)
\(8^{25}=\left(2^3\right)^5=2^{75}\)
Vì \(2^{76}>2^{75}\Rightarrow16^{19}>8^{25}\)
\(b,\)Ta có : \(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
Vì \(125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)
\(M=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{20.21}\)
\(M=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}\right)\)
\(M=3\left(1-\dfrac{1}{21}\right)\)
\(M=3.\dfrac{20}{21}=\dfrac{20}{7}\)
\(N=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{100}{99}\)
\(N=\dfrac{4.9.16.25...100}{3.8.15.24...99}\)
\(N=\dfrac{2.2.3.3.4.4.5.5...10.10}{1.3.2.4.3.5.4.6...9.11}\)
\(N=\dfrac{2.3.4.5...10}{1.2.3...9}.\dfrac{2.3.4.5...10}{3.4.5...11}\)
\(N=10.\dfrac{2}{11}=\dfrac{20}{11}\)
a) \(M=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+......+\dfrac{3}{20.21}\)
= \(3.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+......+\dfrac{1}{20}-\dfrac{1}{21}\right)\)
= \(3.\left(\dfrac{1}{1}-\dfrac{1}{21}\right)\)
= \(3.\dfrac{20}{21}\)
= \(\dfrac{20}{7}\)
b) \(N=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}.......\dfrac{100}{99}\)
= \(\dfrac{4.9.16.25.....100}{3.8.15.24.....99}\)
= \(\dfrac{2.2.3.3.4.4.5.5.......10.10}{1.3.2.4.3.5.4.6......9.11}\)
= \(\dfrac{\left(2.3.4.5.....10\right).\left(2.3.4.5.....10\right)}{\left(1.2.3.4......9\right).\left(3.4.5.....11\right)}\)
= \(\dfrac{10.2}{1.11}\)
= \(\dfrac{20}{11}\)
\(A=\left(\frac{21}{5}-\frac{3}{5}\right):\frac{16}{8}:\frac{9}{10}\)
\(A=\frac{18}{5}:\frac{16}{8}:\frac{9}{10}\)
\(A=\frac{18}{5}\times\frac{8}{16}\times\frac{10}{9}\)
\(A=\frac{9}{5}\times\frac{8}{8}\times\frac{10}{9}\)
\(A=\frac{72}{40}\times\frac{10}{9}\)
\(A=\frac{8}{4}\times\frac{1}{1}\)
\(A=\frac{8}{9}\)
\(B=\left(1\frac{1}{20}+\frac{36}{100}\right):\frac{84}{25}:\left(2+\frac{14}{16}\right)\)
\(B=\left(\frac{21}{20}+\frac{36}{100}\right):\frac{84}{25}:\left(\frac{32}{16}+\frac{14}{16}\right)\)
\(B=\left(\frac{105}{100}+\frac{36}{100}\right):\frac{84}{25}:\frac{46}{16}\)
\(B=\frac{141}{100}:\frac{84}{25}:\frac{46}{16}\)
\(B=\frac{141}{100}\times\frac{25}{84}\times\frac{16}{46}\)
\(B=\frac{141}{4}\times\frac{1}{84}\times\frac{16}{46}\)
\(B=\frac{141}{336}\times\frac{16}{46}\)
\(B=\frac{141}{21}\times\frac{1}{46}\)
\(B=\frac{141}{966}\)
\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{9999}{10000}\\ =\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{99\cdot101}{100\cdot100}\\ =\dfrac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot100\cdot100}\\ =\dfrac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}\\ =\dfrac{1\cdot101}{100\cdot2}\\ =\dfrac{101}{200}\)
\(C=\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\left(1+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{99\cdot101}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{2^2-1}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{3^2-1}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{4^2-1}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{100^2-1}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}\\ =\dfrac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}\\ =\dfrac{100\cdot2}{1\cdot101}=\dfrac{200}{101}\)