A=10+13+.......+79+81
B=1/1.2+1/2.3+1/3.4+.......1/12.20
C=3/7x4/13+3/7+5 4/7
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a ta co ;
13 -12 +11+10-9+8-7-6+5-4+3+2-1
=13-(12-11-10+9) +(8-7-6+5) -(4-3-2+1)
= 13 -0+0 -0
=13
1a) x - 1/3 = 13/6
x = 13/6 + 1/3
x = 5/2
b) -7/5 - x = 13/7 : 13/7
-7/5 - x = 1
x = -7/5 - 1
x = -12/5
2) A = 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/2015.2016
A = 1 -1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2015 - 1/2016
A = 1 - 1/2016
Bài 1: Ta chỉ cần bỏ ngoặc rồi cộng hai phân số để ra kết quả là số tự nhiên là xong
Bài 2:
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+............+\frac{1}{2003.2004}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.............-\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2004}\)
A = \(1-\frac{1}{2004}\)
A = \(\frac{2004}{2004}-\frac{1}{2004}=\frac{2003}{2004}\)
Mk làm mẫu 1 bài nha
Bài 1 :
a, = (1/4+3/4) - (5/13+8/13)+2/11
= 1 - 1 + 2/11
= 2/11
Tk mk nha
c; 17\(\dfrac{2}{31}\) - (\(\dfrac{15}{17}\) + 6\(\dfrac{2}{31}\))
= 17 + \(\dfrac{2}{31}\) - \(\dfrac{15}{17}\) - 6 - \(\dfrac{2}{31}\)
= (17 - 6) - \(\dfrac{15}{17}\) + (\(\dfrac{2}{31}\) - \(\dfrac{2}{31}\))
= 11 - \(\dfrac{15}{17}\)+ 0
= \(\dfrac{172}{17}\)
b; 130\(\dfrac{25}{28}\) + 120\(\dfrac{17}{35}\)
= 130 + \(\dfrac{25}{28}\) + 120 + \(\dfrac{17}{35}\)
= (130 + 120) + (\(\dfrac{25}{28}\) + \(\dfrac{17}{35}\))
= 250 + (\(\dfrac{125}{140}\) + \(\dfrac{68}{140}\))
= 250 + \(\dfrac{193}{140}\)
= 250\(\dfrac{193}{140}\)
ta có;
b=8/3.2/5.3/8.10.19/92
b=16/15.3/8.10.19/92
b=2/5.10.19/92
b=4.19/92
b=19/23
c=-5/7.2/7+-5/7 . 9/14+1/5/7
c=-10/49+(-45)/98+1/5/5
c=131/98
c ) S = 1.2 + 2.3 + 3.4 + .... + 99.100
=> 3S = 1.2.3 + 2.3.3 + 3.4.3 + .... + 99.100.3
=> 3S = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + .... + 99.100.( 101 - 98 )
=> 3S = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 99.100.101 - 98.99.100
=> 3S = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + ( 98.99.100 - 98.99.100 ) + 99.100.101
=> 3S = 99.100.101 => S = \(\frac{99.100.101}{3}\)
d ) Ta có \(\frac{1}{2^2}<\frac{1}{2.1}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
..........
\(\frac{1}{100^2}<\frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1}-\frac{1}{100}=\frac{99}{100}<1\)
Phần A sai nha phải là: 10 + 13 + ... + 79 + 82
đáp án:
SSH: A = ( 82 - 10 ) : 3 + 1 = 25
Tổng: ( 10 + 82 ) . 25 : 2 = 1150
a) A = 10+13 + ...+79 + 81
A = ( 79+10) x 24 : 2 + 81
A = 89 x 24 : 2 +81
A = 1068+ 81
A= 1149
b) chỗ " 1/12.20" phải là 1/12.13 chứ !
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{12.13}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{12}-\frac{1}{13}\)
\(=1-\frac{1}{13}\)
\(=\frac{12}{13}\)
c) \(C=\frac{3}{7}\times\frac{4}{13}+\frac{3}{7}+5\frac{4}{7}\)
\(C=\frac{3}{7}\times\left(\frac{4}{13}+1\right)+\frac{39}{7}\)
\(C=\frac{3}{7}\times\frac{17}{13}+\frac{39}{7}\)
\(C=\frac{51}{91}+\frac{39}{7}\)
\(C=\frac{558}{91}\)