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A=(2/3+3/4+...+99/100)x(1/2+2/3+3/4+...+98/99)-(1/2+2/3+...+99/100)x(2/3+3/4+4/5+...98/99)
ta cho nó dài hơn như sau
A=(2/3+3/4+4/5+5/6+....+98/99+99/100)
ta thấy các mẫu số và tử số giống nhau nên chệt tiêu các số
2:3:4:5...99 vậy ta còn các số 2/100
ta làm vậy với(1/2+2/3+3/4+.....+98/99) thi con 1/99
làm vậy với câu (1/2+2/3+...+99/100) thì ra la 1/100
vậy với (2/3+3/4+...+98/99) ra 2/99
xùy ra ta có 2/100.1/99-1/100.2/99=1/50x1/99-1/100x2/99=tự tinh nhe mình ngủ đây
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: \(=\left(9\cdot\dfrac{8}{2}\right)\cdot25-425=36\cdot25-425=25=5^2\)
a) Cách 1:
\(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}} + \frac{{ - 5}}{{ - 6}}} \right) + \frac{4}{5} = \frac{2}{5} + \frac{5}{6} + \frac{4}{5}\\ = \frac{{12}}{{30}} + \frac{{25}}{{30}} + \frac{{24}}{{30}} = \frac{{61}}{{30}}\end{array}\)
Cách 2:
\(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}} + \frac{{ - 5}}{{ - 6}}} \right) + \frac{4}{5} = \left( {\frac{2}{5} + \frac{4}{5}} \right) + \frac{5}{6}\\ = \frac{6}{5} + \frac{5}{6} = \frac{{36}}{{30}} + \frac{{25}}{{30}} = \frac{{61}}{{30}}\end{array}\)
b) Cách 1:
\(\begin{array}{l}\frac{{ - 3}}{{ - 4}} + \left( {\frac{{11}}{{ - 15}} + \frac{{ - 1}}{2}} \right) = \frac{3}{4} + \frac{{ - 11}}{{15}} + \frac{{ - 1}}{2}\\ = \frac{{45}}{{60}} + \frac{{ - 44}}{{60}} + \frac{{ - 30}}{{60}}\\ = \frac{{ - 29}}{{60}}\end{array}\).
Cách 2:
\(\begin{array}{l}\frac{{ - 3}}{{ - 4}} + \left( {\frac{{11}}{{ - 15}} + \frac{{ - 1}}{2}} \right) = \frac{3}{4} + \frac{{ - 11}}{{15}} + \frac{{ - 1}}{2}\\ = \left( {\frac{3}{4} + \frac{{ - 1}}{2}} \right) + \frac{{ - 11}}{{15}}\\ = \left( {\frac{3}{4} + \frac{{ - 2}}{4}} \right) + \frac{{ - 11}}{{15}}\\ = \frac{1}{4} + \frac{{ - 11}}{{15}}\\ = \frac{{15}}{{60}} + \frac{{ - 44}}{{60}}\\ = \frac{{ - 29}}{{60}}\end{array}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
\(N=\left(2^2+4^2+6^2+...+100^2\right)\left(1^2+3^2+5^2+...+99^2\right)\)
\(N=\left(\frac{100\left(100+1\right)\left(2.100+1\right)}{6}\right)\left(\frac{99\left(2.99-1\right)\left(2.99+1\right)}{3}\right)\)
\(N=338350.1293699=.....\)
\(N=\left(2^2+4^2+....+100^2\right)-\left(1^2+3^2+...+99^2\right)\)
\(=2^2+4^2+6^2+.....+100^2-1^2-3^2-.....-99^2\)
\(=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+......+\left(100-99\right)\left(100+99\right)\)
\(=3+7+....+199\)
\(=3+7+....+197+2\)
\(=4765+2=4767\)