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ta có
A=1/20 + 1/21+1/22+....+1/59
=(1/20+1/21+...+1/39)+(1/40+1//41+....+1/59)<1/20.20+1/40.20=1 + 1/2=3/2
vậy A<3/2
Chúc bạn học tốt nha ^-^
Ta có \(A=\left(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{39}\right)+\left(\frac{1}{40}+\frac{1}{41}+...+\frac{1}{59}\right)\)
\(A< \left(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)+\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)\)
\(A< \frac{20}{20}+\frac{20}{40}\)
\(A< \frac{3}{2}\)
Mẫu số chung : \(LCM\left(60;120;36;90;72\right)=360\)
Quy đồng mẫu số :
\(\dfrac{360}{360}+\dfrac{-6}{360}+\dfrac{57}{360}< \dfrac{10\cdot x}{360}< \dfrac{232}{360}+\dfrac{295}{360}+\dfrac{-6}{360}\)
\(\Leftrightarrow\dfrac{411}{360}< \dfrac{10\cdot x}{360}< \dfrac{521}{360}\)
Vậy tập hợp các giá trị của x là \(x=\left\{42;43;44;45;46;47;48;49;50;51;52\right\}\)
b, Ta có: \(\dfrac{58}{53}>1>\dfrac{36}{55}\)
hay \(\dfrac{58}{53}>\dfrac{36}{55}\)
\(\Rightarrow0-\dfrac{58}{53}< 0-\dfrac{36}{55}\)
\(\Rightarrow\dfrac{-58}{53}< \dfrac{-36}{55}\)
\(1+\frac{-1}{60}+\frac{19}{120}< \frac{x}{36}< \frac{58}{90}+\frac{59}{72}+\frac{-1}{60}\)
=> \(\frac{137}{120}< \frac{x}{36}< \frac{521}{360}\)
=> \(\frac{411}{360}< \frac{10x}{360}< \frac{521}{360}\)
=> 411 < 10x < 521
=> x \(\in\){ 42,43,44,...,52}
\(A=1+4+4^2...+4^{59}\)
\(=\left(1+4+4^2\right)+\left(4^3+4^5+4^6\right)+...+\left(4^{57}+4^{58}+4^{59}\right)\)
\(=21+4^3\cdot21+....+4^{57}\cdot21\)
\(=21\left(1+4^3+4^6+...+4^{57}\right)⋮21\)
\(\Leftrightarrow A⋮21\)
Hok tốt
\(A = 1 + 4 + 4^2 + ... + 4\)\(57\) \(+ 4\)\(58\) \(+ 4\)\(59\)
\(A = ( 1 + 4 + 4^2 ) + ... + ( 4\)\(57\) \(+ 4\)\(58\) \(+ 4\)\(59\)\()\)
\(A = 21 + ... + 4\)\(57\)\(. ( 1 + 4 + 4^2 )\)
\(A = 21 + ... + 4\)\(57\) \(.21\)
\(A = 21 . ( 1 + ... + 4\)\(57\)\()\)\(⋮\)\(21\)
\(Vậy : A \)\(⋮\)\(21\)
\(1+\frac{-1}{60}+\frac{19}{120}<\frac{x}{36}+\frac{-1}{60}<\frac{58}{90}+\frac{59}{72}+\frac{-1}{60}\)
<=>\(\frac{360-6+3.19}{360}<\frac{10x-6}{360}<\frac{58.4+59.5-6}{360}\)
<=>\(\frac{411}{360}<\frac{10x-6}{360}<\frac{467}{360}\)
<=>411<10x-6<467
<=>417<10x<453
<=>41,7<x<45,3
Do x nguyên
=>x={42;43;44;45}
\(T=\frac{1}{59}+\frac{2}{58}+\cdots+\frac{59}{1}\)
\(=\left(\frac{1}{59}+1\right)+\left(\frac{2}{58}+1\right)+\cdots+\left(\frac{58}{2}+1\right)+1\)
\(=\frac{60}{59}+\frac{60}{58}+\cdots+\frac{60}{2}+\frac{60}{60}\)
\(=60\left(\frac12+\frac13+\cdots+\frac{1}{60}\right)\)
=60S
=>\(\frac{S}{T}=\frac{1}{60}\)