Tính \(225.8\). Từ đó suy ra kết quả của :
a) \(\left(-225\right).8\) b) \(\left(-8\right).225\) c) \(8.\left(-224\right)\)
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a. \(160 - \left( {{2^3}{{.5}^2} - 6.25} \right)\)
\(\begin{array}{l} = 160 - \left( {8.25 - 6.25} \right)\\ = 160 - 25.\left( {8 - 6} \right)\\ = 160 - 25.2\\ = 160 - 50\\ = 110\end{array}\)
Ta có: 110 = 2.5.11
b. \(37.3 + 225:{15^2}\)
\(\begin{array}{l} = 37.3 + 225:225\\ = 37.3 + 1\\ = 111 + 1\\ = 112\end{array}\)
Ta có: \(112 = 2^4.7\)
c. \(5871:103 - 64:{2^5}\)
\(\begin{array}{l} = 5871:103 - 64:32\\ = 57 - 2 = 55\end{array}\)
Ta có: 55 = 5. 11
d. \(\left( {1 + 2 + 3 + 4 + 5 + 6 + 7 + 8} \right){.5^2} - 850:2\)
\(\begin{array}{l} = \left[ {\left( {1 + 8} \right) + \left( {2 + 7} \right) + \left( {3 + 6} \right) + \left( {4 + 5} \right)} \right]{.5^2} - 850:2\\ = \left( {9 + 9 + 9 + 9} \right){.5^2} - 850:2\\ = {9.4.5^2} - 850:2\\ = {36.5^2} - 425\\ = {36.5^2} - {5^2}.17\\ = {5^2}.\left( {36 - 17} \right)\\ = {5^2}.19=475\end{array}\)
Ta có: \(475 = 5^2.19\)
a: \(160-\left(2^3\cdot5^2-6\cdot25\right)\)
\(=160-\left(8\cdot25-150\right)\)
\(=160-200+150=10=2\cdot5\)
b: \(=111+225:225=112=2^4\cdot7\)
c: \(=57-64:32=57-2=55=5\cdot11\)
d: \(=\left(9\cdot\dfrac{8}{2}\right)\cdot25-425=36\cdot25-425=25=5^2\)
\(213.3 = 639\)
a) \(\left( { - 213} \right).3 = - \left( {213.3} \right) = - 639\)
b) \(\left( { - 3} \right).213 = - \left( {3.213} \right) = - 639\)
c) \(\left( { - 3} \right).\left( { - 213} \right) = 3.213 = 639\)
a: \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}\cdot\sqrt{\dfrac{49}{4}}\right):\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]:\dfrac{1704}{445}\)
\(=\left(\dfrac{55}{3}:15+\dfrac{26}{3}\cdot\dfrac{7}{4}\right):\left[\left(12+\dfrac{1}{3}+8+\dfrac{6}{7}\right)-\dfrac{7}{18}\right]\cdot\dfrac{445}{1704}\)
\(=\left(\dfrac{55}{45}+\dfrac{91}{6}\right):\left[20+\dfrac{101}{126}\right]\cdot\dfrac{445}{1704}\)
\(=\dfrac{295}{18}:\dfrac{2621}{126}\cdot\dfrac{445}{1704}\)
\(=\dfrac{295}{18}\cdot\dfrac{126}{2621}\cdot\dfrac{445}{1704}\simeq0,21\)
b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
c: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)
\(=\dfrac{1}{n+1}\)
d: \(-66\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124\cdot\left(-37\right)+63\cdot\left(-124\right)\)
\(=-66\cdot\dfrac{33-22+6}{66}+124\left(-37-63\right)\)
\(=-17-12400=-12417\)
e: \(\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)
\(=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(=\dfrac{7}{4}\cdot33\cdot\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)
\(=33\cdot\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(=33\cdot\dfrac{7}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(=33\cdot\dfrac{7}{4}\cdot\dfrac{4}{21}=\dfrac{33\cdot1}{3}=11\)
a) \( - \left( {4 + 7} \right) = - 11\)
\(\begin{array}{l}\left( { - 4 - 7} \right) = \left( { - 4} \right) + \left( { - 7} \right)\\ = - \left( {4 + 7} \right) = - 11\\ \Rightarrow \left( { - 4 - 7} \right) = - \left( {4 + 7} \right)\end{array}\)
b)
\(\begin{array}{l} - \left( {12 - 25} \right) = - \left[ {12 + \left( { - 25} \right)} \right]\\ = - \left[ { - \left( {25 - 12} \right)} \right] = - \left( { - 13} \right) = 13\end{array}\)
\(\begin{array}{l}\left( { - 12 + 25} \right) = 25 - 12 = 13\\ \Rightarrow - \left( {12 - 25} \right) = \left( { - 12 + 25} \right)\end{array}\)
c)
\(\begin{array}{l} - \left( { - 8 + 7} \right) = - \left[ { - \left( {8 - 7} \right)} \right] = - \left( { - 1} \right) = 1\\\left( {8 - 7} \right) = 1\\ \Rightarrow - \left( { - 8 + 7} \right) = \left( {8 - 7} \right)\end{array}\)
d)
\(\begin{array}{l} + \left( { - 15 - 4} \right) = + \left[ {\left( { - 15} \right) + \left( { - 4} \right)} \right]\\ = + \left[ { - \left( {15 + 4} \right)} \right] = + \left( { - 19} \right) = - 19\\\left( { - 15 - 4} \right) = \left( { - 15} \right) + \left( { - 4} \right)\\ = - \left( {15 + 4} \right) = - 19\\ \Rightarrow + \left( { - 15 - 4} \right) = \left( { - 15 - 4} \right)\end{array}\)
e)
\(\begin{array}{l} + \left( {23 - 12} \right) = + 11 = 11\\\left( {23 - 12} \right) = 11\\ \Rightarrow + \left( {23 - 12} \right) = \left( {23 - 12} \right)\end{array}\)
a,-1800
b,-1800
c,-1792