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\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)
\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)
\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)
\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)
Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B
\(\begin{array}{l}\left[ {\left( { - 3} \right) + 4} \right] + 2 = \left( {4 - 3} \right) + 2\\ = 1 + 2 = 3\end{array}\)
\(\begin{array}{l}\left( { - 3} \right) + \left( {4 + 2} \right) = \left( { - 3} \right) + 6\\ = 6 - 3 = 3\end{array}\)
\(\begin{array}{l}\left[ {\left( { - 3} \right) + 2} \right] + 4 = - \left( {3 - 2} \right) + 4\\ = - 1 + 4 = 3\end{array}\)
1) \(\left(\frac{7}{8}-\frac{3}{4}\right).1\frac{1}{3}-\frac{2}{7}.\left(3,5\right)^2\)
= \(\frac{1}{8}.\frac{4}{3}-\frac{2}{7}.\frac{49}{4}\) = \(\frac{-10}{3}\)
2) \(\frac{5}{4}:\frac{1}{2}-\frac{3}{4}\) = \(\frac{7}{4}\)
Ta có
\(A=\frac{\left(1^2-2^2\right)\left(1^2-3^2\right).....\left(1^2-2014^2\right)}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)
\(\Leftrightarrow A=\frac{\left(-1\right)3\left(-2\right)4.....\left(-2013\right)2015}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)
\(\Leftrightarrow A=\frac{\left[\left(-1\right)\left(-2\right)...\left(-2013\right)\right]\left(3.4.5...2015\right)}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)
\(\Leftrightarrow A=\frac{\left(-1\right)2015}{2014.2}=-\frac{2015}{4028}< -\frac{2014}{4028}=-\frac{1}{2}\)
=> A<-1/2
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)..........\left(\frac{1}{2018^2}-1\right)\)
Ta có :
\(\frac{1}{2^2}-1>-\frac{1}{2}\)
\(\frac{1}{3^2}-1>-\frac{1}{2}\)
...........
\(\frac{1}{2018^2}-1>\frac{1}{2}\)
\(\Rightarrow A>B\)
a. \(10,\left(3\right)+0,\left(4\right)-8,\left(6\right)\)
\(\Leftrightarrow\dfrac{31}{3}+\dfrac{4}{9}-\dfrac{26}{3}\)
\(=\dfrac{19}{9}\)
b. \(\dfrac{0,8:\left(\dfrac{4}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(1,08-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(6\dfrac{5}{9}-3\dfrac{1}{4}\right).2\dfrac{2}{17}}+\left(1,2.0,5\right):\dfrac{4}{5}\)
\(=\dfrac{0,8}{0,6}+\dfrac{1,75}{7}+0,6:\dfrac{4}{5}\)
\(=\dfrac{0,8}{0,6}+\dfrac{1,75}{7}+\dfrac{3}{4}\)
\(=\dfrac{7}{3}\)
Giải
a, Ta có :
\(25^2=625\text{ }\Rightarrow\text{ }1,25^2=1,625\)
\((1,25)^2=1,25\times1,25=1,5625\)
Vì \(1,625\ne1,5625\Rightarrow1,25^2\ne\left(1,25\right)^2\).
b, Ta có :
\(\left(2^2\right)^{^3}=2^{2\times3}=2^6=64\)
\(2^{2^3}=2^8=256\)
Vì \(64\ne256\Rightarrow\left(2^2\right)^{^3}\ne2^{2^3}\).
* Hãy cẩn thận khi viết luỹ thừa của số thập phân và luỹ thừa của luỹ thừa ! *
a ) \(1,25^2=1,5625\)
\(\left(1,25\right)^2=1,5625\)
= > 1,25 2 = ( 1,25 )2
b) \(\left(2^2\right)^3=64\)
\(2^{2^3}=256\)
= > \(2^{2^3}>\left(2^2\right)^3\)