tính 1/4+1/12+1/24+...+1/180
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
=1/2(1/2+1/6+1/12+1/20+...+1/90)
=1/2(1-1/2+1/2-1/3+...+1/9-1/10)
=1/2*9/10=9/20
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, (-25) + 276 - (276 - 25)
= (-25) + 276 - 276 + 25
= [ (-25) + 25] - (276 - 276)
= 0 - 0
= 0
b, 24 x 46 + 24 x 53 + 24
= 24 x ( 46 + 53 + 1)
= 24 x 100
= 2400
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. Cho phân số a/b, b>a. C/m :
a) Nếu a/b < 1 thì a+1/b+1 > a/b
b) Nếu a/b > 1 thì a/b > a+1/b+1
2. Tính :
3/4 + 3/12 + 3/24 + 3/40 + 3/60 + 3/84 + 3/112 + 3/144 + 1/180
Toán lớp 6
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1/4 x 8 + 1/8 x 12 + 1/12 x 16 + ... + 1/176 x 180
=> 4A = 4/4 x 8 + 4/8 x 12 + 4/12 x 16 + ... + 4/176 x 180
=> 4A = 1/4 - 1/8 + 1/8 - 1/12 + 1/12 - 1/16 + ... 1/176 - 1/180
=> 4A = 1/4 - 1/180
=> 4A = 45/180 - 1/180
=> 4A = 44/180
=> 4A = 11/45
=> A = 11/45 : 4
=> A = 11/180
Vậy A = 11/180
A = \(\dfrac{1}{4\times8}\) + \(\dfrac{1}{8\times12}\) + \(\dfrac{1}{12\times16}\) +...+ \(\dfrac{1}{176\times180}\)
A = \(\dfrac{1}{4}\) \(\times\)( \(\dfrac{4}{4\times8}\)+ \(\dfrac{4}{12\times16}\)+...+ \(\dfrac{4}{176\times180}\))
A = \(\dfrac{1}{4}\) \(\times\)( \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{16}\) +...+ \(\dfrac{1}{176}\) - \(\dfrac{1}{180}\))
A = \(\dfrac{1}{4}\) \(\times\)(\(\dfrac{1}{4}\) - \(\dfrac{1}{180}\))
A = \(\dfrac{1}{4}\) \(\times\)\(\dfrac{11}{45}\)
A = \(\dfrac{11}{180}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2A=1/2+1/6+1/12+1/20+1/30+1/42
=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
=1-1/7
=6/7
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(\Leftrightarrow x=UCLN\left(24;36;150\right)=6\)
2: \(\Leftrightarrow x\in\left\{24;48;72;...\right\}\)
mà 16<=x<=50
nên \(x\in\left\{24;48\right\}\)
3: \(\Leftrightarrow x\inƯ\left(6\right)\)
mà x>-10
nên \(x\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
4: \(\Leftrightarrow x\in BC\left(4;5;8\right)\)
\(\Leftrightarrow x\in\left\{...;-40;0;40;80;120;160;200;...\right\}\)
mà -20<x<180
nên \(x\in\left\{0;40;80;120;160\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Cho tg tren la A
A=\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)
\(A=2\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}+\frac{1}{12.14}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{14}\right)\)
\(A=2.\frac{3}{7}\)
\(A=\frac{6}{7}\)
\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{180}=\frac{1}{2}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)=\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)\(=\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{2}.\left(1-\frac{1}{10}\right)=\frac{1}{2}.\frac{9}{10}=\frac{9}{20}\)