Các câu hỏi tương tự
Bài 1:tính tích
a, A=(1-1/2).(1-1/3).(1-1/4)...(1.1/999).(1-1/1000)
b, B= 3/4.8/9.1/16...2499/2500
Tính các tích sau:
A= 3/4.8/9.15/16...9999/10000
B=(1-1/4).(1-1/9)...(1-1/10000).
C=(1+1/1.3)(1+1/2.4)(1+1/3.15)....(1+1/99.100).
Chứng tỏ :a, A dfrac{1}{2.4}+dfrac{1}{4.6}+...+dfrac{1}{2022.2024} dfrac{1}{4}b, B dfrac{1}{1.3}+dfrac{1}{3.5}+...+dfrac{1}{2013.2015} dfrac{1}{2}c, C dfrac{1}{3^2}+dfrac{1}{5^2}+dfrac{1}{7^2}+...+dfrac{1}{2013^2} dfrac{1}{4}d, D dfrac{1}{2^2}+dfrac{1}{4^2}+dfrac{1}{6^2}+...+dfrac{1}{2014^2} dfrac{1}{2}
Đọc tiếp
Chứng tỏ :
a, A = \(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{2022.2024}\) < \(\dfrac{1}{4}\)
b, B =\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2013.2015}< \dfrac{1}{2}\)
c, C =\(\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2013^2}< \dfrac{1}{4}\)
d, D =\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2014^2}< \dfrac{1}{2}\)
Tìm x€ Z, biết:
1) 7/11 x -3/7< x< -1/5 :1/20 -(-2)3
2) tính tổng S= 1/1.3 + 1/2.4 + 1/3.5 +...+1/7.9 + 1/8.10
Cách làm
B1:Tính nhanh:
a)M= -1/3 . 141/17 - 39/3 . -1/17
b)N= -9/16 . 13/3 - (-3/4)^2 . 19/3
c)P=(1+1/1.3) (1+1/2.4) (1+1/3.5)......(1+1/99.101)
B2:Tìm x,biết
a)1/2+3/2:x=1/4
b)3/4+1/4.x=7
c)1/2.4 + 1/4.6 +.........+1/(2x-2).2x = 11/48
B3:Tìm x,biết
a)(x-1/2)^2=1/81
b)x+x(1+1/x)+x(1+2/x)=1/3
c)2/x+4=3/x+5
Giải hộ mk cái . mk đang cần gấp. ~_~
Bài 1: tính
A=1+7+7^2 +7^3+..+7^2007
B=1+4+4^2+4^3+...+4^100
C=1+3^2+3^4+3^6+3^8+...+3^100
D=7+7^3+7^5+7^7+7^9+...+7^99
E=2+2^3+2^5+2^7+2^9+...+2^9009
G=1+2^2+2^4+2^8+2^10+...+2^200
Bài 2:tính
A=1^2+3^2+5^3+7^3+...+99^3
B=11^2+13^2+15^2+...+2009^2
C=1.2+2.3+3.4+4.5+...+99.100
d=20^2+22^2+...+48^2+50^2
e=1+4+9+16+25+36+...+1000
bài3
A=1.2+3.5+5.7+...+97.99
1) Tính
A= 3/4.8/9.15/16...9999/10000
B= (1-1/4).(1-1/9)...(1-1/1000)
A=1/1×2+1/3×4+1/4×5+...1/999×1000
B=1/501×1000+1/502×999+...+1/999×502+1/1000×501
Tính A/B
Tính tích:
B= (1- 1/4).(1- 1/9).... (1- 1/1000)
C= 3/4 . 8/9 . 15/16 .... 99/100
D= (1+ 1/1.3) . (1+ 1/2.4) . (1+ 1/3.5).... (1+ 1/99.100)
Ai nhanh mình tích! ^_^