6^3×5^2×(5/6)^3
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\(\dfrac{2002\times2003+4004}{2005\times2004-4010}\\ =\dfrac{2002\times2003+2002\times2}{2005\times2004-2005\times2}\\ =\dfrac{2002\times2005}{2005\times2002}\\ =1\)
(\(\dfrac{1}{9}\))2 = (\(\dfrac{1^2}{3^2}\))2= ((\(\dfrac{1}{3}\))2)2= (\(\dfrac{1}{3}\))4
`@` `\text {Ans}`
`\downarrow`
a)
\(2\dfrac{3}{4}-a+\dfrac{1}{4}=1\dfrac{1}{2}\)
`\Rightarrow`\(\dfrac{11}{4}-a+\dfrac{1}{4}=\dfrac{3}{2}\)
`\Rightarrow`\(\dfrac{10}{4}-a=\dfrac{3}{2}\)
`\Rightarrow`\(a=\dfrac{10}{4}-\dfrac{3}{2}\)
`\Rightarrow`\(a=\dfrac{5}{2}-\dfrac{3}{2}\)
`\Rightarrow`\(a=\dfrac{2}{2}=1\)
Vậy, `a=1`
b)
\(3\dfrac{1}{4}-a-1\dfrac{3}{4}=\dfrac{7}{9}\)
`\Rightarrow`\(a+\dfrac{3}{2}=\dfrac{7}{9}\)
`\Rightarrow`\(a=\dfrac{7}{9}-\dfrac{3}{2}\)
`\Rightarrow`\(a=-\dfrac{13}{18}\)
Vậy, \(a=-\dfrac{13}{18}\)
c)
\(2\dfrac{5}{6}-1\dfrac{1}{2}-a=\dfrac{1}{6}\)
`\Rightarrow`\(\dfrac{4}{3}-a=\dfrac{1}{6}\)
`\Rightarrow`\(a=\dfrac{4}{3}-\dfrac{1}{6}\)
`\Rightarrow`\(a=\dfrac{7}{6}\)
Vậy, \(a=\dfrac{7}{6}\)
\(A=\dfrac{1}{1\cdot2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4\cdot5}+\dfrac{1}{3\cdot4\cdot5\cdot6}+....+\dfrac{1}{9\cdot10\cdot11\cdot12}\)
\(3A=\dfrac{3}{1\cdot2\cdot3\cdot4}+\dfrac{3}{2\cdot3\cdot4\cdot5}+\dfrac{3}{3\cdot4\cdot5\cdot6}+...+\dfrac{3}{9\cdot10\cdot11\cdot12}\)
\(3A=\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4}-\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{9\cdot10\cdot11}-\dfrac{1}{10\cdot11\cdot12}\)\(3A=\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{10\cdot11\cdot12}\)
\(A=\dfrac{1}{2}-\dfrac{1}{440}\)
\(A=\dfrac{219}{440}\)
63x52x(\(\dfrac{5}{6}\))3=52x53=55.