3/4 so sánh với 9/2
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{9}{10!}\)
\(A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{10-1}{10!}\)
\(A=\left(\frac{2}{2!}-\frac{1}{2!}\right)+\left(\frac{3}{3!}-\frac{1}{3!}\right)+\left(\frac{4}{4!}-\frac{1}{4!}\right)+...+\left(\frac{10}{10!}-\frac{1}{10!}\right)\)
\(A=\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+...+\left(\frac{1}{9!}-\frac{1}{10!}\right)\)
\(A=1-\frac{1}{10!}< 1\)
vậy A < 1 vì \(0< \frac{1}{10!}< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
M = \(\frac{1}{2}\)+ \(\frac{2}{3}\)+ \(\frac{3}{4}\)+\(\frac{4}{5}\)+ \(\frac{5}{6}\)+ \(\frac{6}{7}\)+ \(\frac{7}{8}\)+ \(\frac{8}{9}\)+ \(\frac{9}{10}\)= \(\frac{17819}{2520}\)
Vậy: M > 1
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(\dfrac{3}{4}\) + \(\dfrac{3}{9}\) + \(\dfrac{3}{16}\) + \(\dfrac{3}{25}\) +..............+ \(\dfrac{3}{(3n)^2}\)
A = ( \(\dfrac{3}{4}\) + \(\dfrac{3}{9}\) + \(\dfrac{3}{16}\)+ \(\dfrac{3}{25}\)) +.....+ \(\dfrac{3}{(3n)^2}\)
A = 3. ( \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{5^2}\))+............+ \(\dfrac{3}{(3n)^2}\)
A = 3.( \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + \(\dfrac{1}{4.4}\) + \(\dfrac{1}{5.5}\)) +............+ \(\dfrac{3}{(3n)^2}\)
Vì \(\dfrac{1}{2}\) > \(\dfrac{1}{3}\) > \(\dfrac{1}{4}\) > \(\dfrac{1}{5}\)Ta có : \(\dfrac{1}{2.2}>\dfrac{1}{2.3}>\dfrac{1}{3.3}>\dfrac{1}{3.4}>\dfrac{1}{4.4}>\dfrac{1}{4.5}>\dfrac{1}{5.5}>\dfrac{1}{5.6}\)
A > 3. ( \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\)) + ............+ \(\dfrac{1}{(3n)^2}\)
A > 3. ( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)) +.....+ \(\dfrac{1}{(3n)^2}\)
A > 3.( \(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)) +..............+ \(\dfrac{1}{(3n)^2}\)
A > 3. \(\dfrac{1}{3}\) +...............+ \(\dfrac{1}{(3n)^2}\)
A > 1 +..........+ \(\dfrac{1}{9n^2}\) > 1
A > 1
![](https://rs.olm.vn/images/avt/0.png?1311)
DẶT A= BIỂU THỨC TRÊN
A=2+1+1+..+1-(1/4+1/9+...+1/2500)
ĐẶT S=1/4+1/9+...+1/2500
S=1/2^2+1/3^2+...+1/50^2
SÓ SỐ HẠNG CỦA S:
(50-2)/1+1=49
SUY RA
1+1+...+1=49
SUY RA A=2+49-S
A=51-S
TAO CÓ :
S<1/1.2+1/2.3+...+1/49.100
S<1-1/2+1/2-1/3+...+1/49-1/50
S<1-1/50
S<49/50
SUY RA A>51-49/50
SUY RA A>50
![](https://rs.olm.vn/images/avt/0.png?1311)
1 + 2 + 1 + 3 + 1 + 4 + .. + 1 + 100
= 99 x 1 + ( 2 + 3 + 4 + ... + 100 )
= 5148 ( 1 )
1 + 3 + 5 +7 + ... + 301
= \(\frac{\left[\left(301-1\right):2+1\right].\left(301+1\right)}{2}\)
= 22801 (2)
Từ ( 1) và (2) => 1+3+ ....+ 301 > 1+2+1+3+1+4 +...+ 1 + 100
b) làm tương tự
Câu đầu bé theo linh cảm thôi
Câu hai:Lớn vì phép đầu với phép hai ko có số 1,25 là bằng nhau nhưng lại có.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1>\frac{2}{3}\)
\(1>\frac{3}{4}\)
\(1=\frac{5}{5}\)
\(1< \frac{9}{8}\)
\(1< \frac{12}{2}\)
\(1< \frac{14}{13}\)
Ta có:\(\frac{9}{2}=\frac{9x2}{2x2}=\frac{18}{4}\)
Vì \(\frac{3}{4}< \frac{18}{4}\Rightarrow\frac{3}{4}< \frac{9}{2}\)
Học tốt.
Ta có :
\(\frac{9}{2}=\frac{9\times2}{2\times2}=\frac{18}{4}\)
\(\frac{3}{4}< \frac{18}{4}\Rightarrow\frac{3}{4}< \frac{9}{2}\)
#Hok tốt