Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a: \(\dfrac{25}{42}-\dfrac{20}{63}=\dfrac{75-40}{126}=\dfrac{35}{126}=\dfrac{5}{18}\)
b: \(\dfrac{9}{20}-\dfrac{13}{75}-\dfrac{1}{6}=\dfrac{135}{300}-\dfrac{52}{300}-\dfrac{50}{300}=\dfrac{33}{300}=\dfrac{11}{100}\)
a: \(A=\dfrac{5}{7}-\dfrac{2}{7}+\dfrac{8}{11}+\dfrac{3}{11}+\dfrac{1}{2}=\dfrac{3}{7}+\dfrac{1}{2}+1=\dfrac{6+7+14}{14}=\dfrac{27}{14}\)
b: \(B=\dfrac{11}{17}+\dfrac{6}{17}-\dfrac{8}{19}-\dfrac{30}{19}+\dfrac{-3}{4}=1-2-\dfrac{3}{4}=-1-\dfrac{3}{4}=-\dfrac{7}{4}\)
c: \(C=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}=\dfrac{49}{50}\)
a ) 15 : ( x + 2 ) = 3
x + 2 = 15 : 3
x + 2 = 5
x = 5 - 2
x = 3
b ) 541 - ( 218 + x ) = 73
218 + x = 541 - 73
218 + x = 468
x = 468 - 218
x = 250
c ) 20 : ( x + 1 ) = 2
x + 1 = 20 : 2
x + 1 = 10
x = 10 - 1
x = 9
d ) 96 - 3 . ( x + 1 ) = 42
3 . ( x + 1 ) = 96 - 42
3 . ( x + 1 ) = 54
x + 1 = 54 : 3
x + 1 = 18
x = 18 - 1
x = 17
Chuỗi đã cho là chuỗi vô hạn
k = n
k ^ k = 1 ^ 1 + 2 ^ 2 + 3 ^ 3 + Sự ..n ^ n (1)
k = 1
Chúng ta có thể viết như sau,
k = n
∑k ^ k = 1 ^ 1 + 2 ^ 2 + 3 ^ 3 + Cáp. (N-1) ^ (n-1) + n ^ n (2)
k = 1
Trừ n ^ n ở cả hai bên
k = n
∑k ^ kn ^ n = 1 ^ 1 + 2 ^ 2 + 3 ^ 3 + '. (N-1) ^ (n-1) (2)
k = 1
k = n-1
∑k ^ k- (n ^ n) = x
k = 1
a: \(100+98+96+...+2-97-95-...-1\)
\(=100+\left(98-97\right)+\left(96-95\right)+...+\left(2-1\right)\)
\(=100+1+...+1\)
\(=100+49=149\)
b: \(1+2-3-4+5+6-7-8+9+10-11-12+...+...-299-300+301+302\)
\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(297+298-299-300\right)+603\)
\(=603+\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)
\(=603-4\cdot75=603-300=303\)
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2018}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2018.2019}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2018.2019}\right)\)
\(=2\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2019-2018}{2018.2019}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2019}\right)\)
\(=\frac{2017}{2019}\)
1+12(1+2)+13(1+2+3)+...+120(1+2+...+20)B=1+12(1+2)+13(1+2+3)+...+120(1+2+...+20)
1+12.2.3:2+13.3.4:2+...+120.20.21:2B=1+12.2.3:2+13.3.4:2+...+120.20.21:2
=22+32+...+212B=22+32+...+212
=2+3+...+212B=2+3+...+212
=2302B=2302
=115
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}+\left(1+2+3\right)+...+\frac{1}{20}+\left(1+2+...+20\right)\)
= \(1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+...+\frac{1}{20}.20.21:2\)
= \(\frac{2}{2}+\frac{3}{2}+...+\frac{21}{2}\)
=\(\frac{2+3+...+21}{2}\)
= \(\frac{230}{2}=115\)
Học tốt!