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A=\(\frac{1}{30}\)+\(\frac{1}{42}\)+\(\frac{1}{56}\)+\(\frac{1}{72}\)+\(\frac{1}{90}\)+\(\frac{1}{110}\)+\(\frac{1}{132}\)
A=\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)+\(\frac{1}{8.9}\)+\(\frac{1}{9.10}\)+\(\frac{1}{10.11}\)+\(\frac{1}{11.12}\)
A= \(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{8}\)+\(\frac{1}{8}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{10}\)+\(\frac{1}{10}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{12}\)
A= \(\frac{1}{5}\)-\(\frac{1}{12}\)=\(\frac{7}{60}\)
Câu 2:
b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}\)
c: \(\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{110}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=\dfrac{1}{4}-\dfrac{1}{11}=\dfrac{7}{44}\)
a: =(-2)+(-2)+...+(-2)=-2x10=-20
b: =(-3)+(-3)+...+(-3)+103=-3x50+103=-150+103=-47
a) 1 + (-3) + 5 + (-7) + ... + 17 + (-19)
= (-2) + (-2) + ... + (-2) (có 5 số -2)
= (-2) . 5
= -10
b) 1 - 4 + 7 - 10 + ... - 100 + 103
= (-3) + (-3) + ... + (-3) + 103 (có 17 số -3)
= (-3) . 17 + 103
= -51 + 103
= 52
c) 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... - 99 - 100 + 101 + 102
= 1 + (2 - 3 - 4 + 5) + (6 - 7 - 8 + 9) + ... + (98 - 99 - 100 + 101) + 102
= 1 + 0 + 0 + ... + 0 + 102
= 103
a) 1 + (-3) + 5 + (-7) + ... + 17 + (-19)
= -2 + (-2) + (-2) + (-2) + (-2)
= -2.5
= -10
b) 1 - 4 + 7 - 10 + ... - 100 + 103
= -3 + (-3) + ... + (-3) + 103 (16 số (-3))
= -16.3 + 103
= -48 + 103
= 55
c) 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... - 99 - 100 + 101 + 102
= -4 + (-4) + ... + (-4) + 101 + 102 (25 số (-4))
= -4.25 + 203
= -100 + 203
= 103
ai giúp mk cũng tick hết á!!!
b1
a) \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{5}-\dfrac{1}{10}\)
\(=\dfrac{2}{10}-\dfrac{1}{10}\)
\(=\dfrac{1}{10}\)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{1}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
c) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)
\(=\dfrac{1}{3}-\dfrac{1}{11}\)
\(=\dfrac{8}{33}\)
d) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=\dfrac{1}{3}-\dfrac{1}{101}\)
\(=\dfrac{98}{303}\)
Câu 1:Để (n^2-4)*(n^2-25)<0
Thì n^2-4 và n^2-25 phải khác dấu
Mà n^2-4>n^2-25
=>n^2-4>0
n^2-25<0
=>n^2>4=>n>2(1)
=>n^2<25=>n<5(2)
Từ (1) và (2)=>2<n<5
=>n=3 hoặc n=4
Câu 2:Ta có 1/1^2=1
1/2^2<1/1*2
1/3^2<1/2*3
...
1/50^2<1/49*1/50
=>1/1^2+1/2^2+1/3^2+...+1/50^2<1+1/1*2+1/2*3+...+1/49*50
1/1^2+1/2^2+1/3^2+...+1/50^2<1+1-1/2+1/2-1/3+...+1/49-1/50
1/1^2+1/2^2+1/3^2+...+1/50^2<2-1/50<2
=>1/1^2+1/2^2+1/3^2+...+1/50^2<2
\(S=3+\frac{3}{2}+\frac{3}{2^2}+....+\frac{3}{2^9}\)
=>\(2S=6+3+\frac{3}{2}+....+\frac{3}{2^8}\)
=>\(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+....+\frac{3}{2^9}\right)\)
=>\(S=6-\frac{3}{2^9}=\frac{3069}{512}\)
Gọi tử số của B là a và mẫu là b
\(a=1+2+2^2+2^3+...+2^{2008}\)
\(2a=2+2^2+2^3+...+2^{2009}\)
\(2a-a=\left(2+2^2+2^3+...+2^{2009}\right)-\left(1+2+2^2+2^3+...+2^{2008}\right)\)
\(a=2^{2009}-1\)
\(a=\frac{2^{2009}-1}{1-2^{2009}}\)
\(a=1\)
$2a-a=\left(2+2^2+2^3+...+2^{2009}\right)-\left(1+2+2^2+...+2^{2008}\right)$2a−a=(2+22+23+...+22009)−(1+2+22+...+22008)
$a=\left(2-2\right)+\left(2^2-2^2\right)+...+\left(2^{2008}-2^{2008}\right)+2^{2009}-1$a=(2−2)+(22−22)+...+(22008−22008)+22009−1
$a=0+0+0+2^{2009}-1$a=0+0+0+22009−1
$a=2^{2009}-1$a=22009−1
$B=\frac{2^{2009}-1}{1-2^{2009}}$B=22009−11−22009
B= -1