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K = (\(\frac{3^5}{3}+\frac{3^5}{3^2}+\frac{3^5}{3^3}+\frac{3^5}{3^4}\))+...+\(\left(\frac{3^{101}}{3^{97}}+\frac{3^{101}}{3^{98}}+\frac{3^{101}}{3^{99}}+\frac{3^{101}}{3^{100}}\right)\)
\(=\left(3^1+3^2+3^3+3^4\right)+...+\left(3^1+3^2+3^3+3^4\right)\)
\(=120+...+120\)(Có 25 số 120)
\(=25.120\)
\(=300\)
vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
S = 101 + (-102) + 103 + (-104) + ... + 2017 + (-2018)
Khi số âm là số nguyên, ta có số số hạng là:
(2018 - 101) : 1 + 1 = 1918 (số hạng)
S = [101 + (-102)] + [103 + (-104)] + ... + [2017 + (-2018)]
S = (- 1) + (-1) + ... + (-1)
Có số số hạng là:
1918 : 2 = 959 (số hạng)
S = (-1) \(\times\) 959
S = - 959
P=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)
=0+0+...+0
=0
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Tách đôi ra ÁP CÔNG THÚC TỔNG VÀO:
a=1+5+..+97
B=-3+-7+..-99=-(3+...+99)
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Mk lm đc câu a thôi nhé !
A= 150-(100-99+98-97+...-3+2-1)
từ 1-100 có 100 SH. Ta nhóm 4 số vs nhau như sau : (100-00+98-87)+(...)+(4-3+2-1)
Có tất cả số nhóm là : 100:4=25 nhóm. Mà mỗi nhóm ta tính có kết quả là 2, vậy tao có
A=150-(2.25)
A=150-50
A=100
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{53}{101}.\frac{-13}{97}+\frac{53}{101}.\frac{-84}{97}\)
\(=\frac{53}{101}\left(\frac{-13}{97}+\frac{-84}{97}\right)\)
\(=\frac{53}{101}.\frac{-97}{97}\)
\(=\frac{53}{101}.\left(-1\right)\)
\(=\frac{-53}{101}\)
b) \(\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{57}-\frac{1}{5757}\right).0\)
\(=0\)
c) \(\frac{3^2}{25}.\frac{75}{-21}.\frac{50}{35}\)
\(=\frac{3^2.75.50}{25.\left(-21\right).35}\)
\(=\frac{3.3.25.3.5.5.2}{25.3.\left(-7\right).5.7}\)
\(=\frac{3.3.5.2}{\left(-7\right).7}\)
\(=\frac{90}{-49}\)
d) \(\frac{25.48-25.18}{20.5^3}\)
\(=\frac{25\left(48-18\right)}{10.2.125}\)
\(=\frac{25.10.3}{10.2.25.5}\)
\(=\frac{3}{10}\)
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\(T=\left(\frac{1}{2}-\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{5}\right)\left(\frac{1}{2}-\frac{1}{7}\right).....\left(\frac{1}{2}-\frac{1}{99}\right)\)
\(T=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-........-\frac{1}{99}\right)\)
\(T=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-..........-\frac{1}{99}\right)\)
\(T=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(T=\frac{1}{2}\left(\frac{99}{297}-\frac{3}{297}\right)\)
\(T=\frac{1}{2}.\frac{96}{297}\)
\(T=\frac{1}{2}.\frac{32}{99}\)
\(T=\frac{16}{99}\)
\(T=\left(\frac{1}{2}-\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{5}\right)\left(\frac{1}{2}-\frac{1}{7}\right)\cdot.....\cdot\left(\frac{1}{2}-\frac{1}{99}\right)\)
\(=\frac{1}{2\cdot3}\cdot\frac{3}{2\cdot5}\cdot\frac{5}{2\cdot7}\cdot.....\cdot\frac{97}{2\cdot99}\)
\(=\frac{1\cdot3\cdot5\cdot.....\cdot97}{2^{49}\left(3\cdot5\cdot7\cdot....\cdot99\right)}=\frac{1}{2^{49}\cdot99}\)
\(B=\left[1+\left(-3\right)\right]+\left[5+\left(-7\right)\right]+........+\left[97+\left(-99\right)\right]+101\)
\(=\left(-2\right)+\left(-2\right)+.......+\left(-2\right)+101\)( có 25 số -2)
\(=\left(-2\right).25+101\)
\(=\left(-50\right)+101\)
\(=51\)
\(B=1+\left(-3\right)+5+\left(-7\right)+...+97+\left(-99\right)+101\)
\(\Rightarrow B=\left[1+\left(-3\right)\right]+\left[5+\left(-7\right)\right]+...+\left[97+\left(-99\right)\right]+101\)
\(\Rightarrow B=\left(-2\right)+\left(-2\right)+...+\left(-2\right)+101\)
có 25 số -2
\(\Rightarrow B=\left(-2\right).25+101\)
\(\Rightarrow B=-50+101\)
\(\Rightarrow B=51\)