Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{120}\left(a\right)\)
\(\Rightarrow A=\left(\dfrac{1}{101}+\dfrac{1}{102}+...\dfrac{1}{125}\right)+\left(\dfrac{1}{126}+\dfrac{1}{127}+...\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...\dfrac{1}{175}\right)+\left(\dfrac{1}{176}+\dfrac{1}{177}+...\dfrac{1}{200}\right)\)
\(\Rightarrow A>25.\dfrac{1}{125}+25.\dfrac{1}{150}+25.\dfrac{1}{175}+25.\dfrac{1}{200}\)
\(\Rightarrow A>\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\)
\(\Rightarrow A>\dfrac{168+140+120+105}{840}=\dfrac{533}{840}>\dfrac{5}{8}\left(\dfrac{533}{840}>\dfrac{525}{840}\right)\)
\(\Rightarrow A>\dfrac{5}{8}\left(1\right)\)
\(\left(a\right)\Rightarrow A=\left(\dfrac{1}{101}+...\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+...\dfrac{1}{140}\right)+\left(\dfrac{1}{141}+...\dfrac{1}{160}\right)+\left(\dfrac{1}{161}+...\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+...\dfrac{1}{200}\right)\)
\(\Rightarrow A< 20.\dfrac{1}{100}+20.\dfrac{1}{120}+20.\dfrac{1}{140}+20.\dfrac{1}{160}+20.\dfrac{1}{180}\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{504+420+360+315+280}{2520}=\dfrac{1879}{2520}< \dfrac{3}{4}\left(\dfrac{1879}{2520}< \dfrac{1890}{2520}\right)\)
\(\Rightarrow A< \dfrac{3}{4}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{5}{8}< A< \dfrac{3}{4}\left(dpcm\right)\)
Ta có : \(\frac{1}{101}\) > \(\frac{1}{150}\)
\(\frac{1}{102}\) > \(\frac{1}{150}\)
.....................................................
\(\frac{1}{149}\) > \(\frac{1}{150}\)
=> \(\frac{1}{101}\) + \(\frac{1}{102}\) + .......... + \(\frac{1}{150}\) > \(\frac{1}{150}\) + \(\frac{1}{150}\) + .......... + \(\frac{1}{150}\)( có 50 p/s ) = \(\frac{1}{150}\) . 50 = \(\frac{1}{3}\)(1)
Ta lại có : \(\frac{1}{151}\) > \(\frac{1}{200}\)
\(\frac{1}{152}\) > \(\frac{1}{200}\)
............................................
\(\frac{1}{199}\)> \(\frac{1}{200}\)
=> \(\frac{1}{151}\) + \(\frac{1}{152}\) + .................. + \(\frac{1}{200}\) > \(\frac{1}{200}\)+ \(\frac{1}{200}\) + ...................+ \(\frac{1}{200}\)(có 50 p/ )=\(\frac{1}{200}\) . 50 = \(\frac{1}{4}\)(2)
Từ (1) và (2)
=> \(\frac{1}{101}\)+ \(\frac{1}{102}\) + \(\frac{1}{103}\) + ...................+ \(\frac{1}{200}\)> \(\frac{1}{3}\) + \(\frac{1}{4}\) = \(\frac{4}{12}\) + \(\frac{3}{12}\) = \(\frac{7}{12}\)
Vậy A > \(\frac{7}{12}\)
Ta có :\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
=> Đpcm
a) \(5^5-5^4+5^3\)
\(=5^3.5^2-5^3.5+5^3\)
=\(5^3.\left(5^2-5+1\right)\)
\(=5^3.21\)
vì \(21⋮7\Rightarrow5^3.21⋮7\)
\(\Rightarrow5^5-5^4+5^3⋮7\)
a)
55-54+53
=53(52-5+1)
=53.21
=53.3.7
vì trong tích 53.3.7 có chứa 1 thừa số chia hết cho 7
=> 53.3.7 chia hết cho 7 hay 55-54+53 chia hết cho 7
b)
76+75-74
=74(72+7-1)
=74.55
=74.5.11
vì trong tích 74.5.11 có 1 thừa số chia hết cho 11 nên
74.5.11 chia hết cho 11 hay 76+75-74 chia hết cho 11