Tính: A=1+1/2(1+2)+1/3(1+2+3)+....+1/16(1+2+3+....+16)
NL
Những câu hỏi liên quan
Tính nhanh: A = 1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+....+1/16(1+2+3+...+16)
Ai nhanh mk tích
A, thực hiện phép tính
M=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/16(1+2+3+4+...+16)
B, tìm x biết
|4x+3|-|x-1|=7
Tính A=3.[2^2+1].[2^4+1].[2^8+1].[2^16+1]
: A = 3(2²+1)(2^4 + 1)....(2^64 + 1) + 1
= (2²-1)(2²+1)(2^4 + 1)....(2^64 + 1) + 1
= (2^4 - 1)(2^4 + 1)....(2^64 + 1) + 1
= (2^8 - 1).(2^8 + 1)(2^16 + 1)(2^32 + 1)(2^64 + 1) + 1
= (2^16 - 1)(2^16 + 1)(2^32 + 1)(2^64 + 1) + 1
= (2^32 - 1)(2^32 + 1)(2^64 + 1) + 1
= (2^64 - 1)(2^64 + 1) + 1 = 2^128 - 1 + 1 = 2^128.
Đúng 0
Bình luận (0)
tính \(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\)
\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\\ P=1+\dfrac{1}{2}.\dfrac{2.3}{2}+\dfrac{1}{3}.\dfrac{3.4}{2}+...+\dfrac{1}{16}.\dfrac{16.17}{2}\\ P=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{17}{2}\\ P=\dfrac{1}{2}\left(2+3+4+...+17\right)\\ P=\dfrac{1}{2}\left(\dfrac{17.18}{2}-1\right)\\ P=\dfrac{1}{2}.152=76\)
Đúng 0
Bình luận (5)
M= 1+\(\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.......+\frac{1}{16}\left(1+2+3+...+16\right)\)
Tính M
tính giá trị của biểu thức sau
a, 1/3-1/7-1/7 /2/3-2/7-2/13 ✖3/4-3/16-3/64-3/256 / 1-1/4-1/16-1/64 +5/8
giúp mink vs nha mn
Tính nhanh:D=(3+1)x(3^2+1)x(3^4+1)x(3^8+1)x(3^16+1)
rút gọn biểu thức
a) A=16^8 -1/(2+1)(2^2+1)(2^4+1)(2^8+1(3^16+1)
b) B=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)/9^16-1
giúp mk vs ah mk đang cần gấp ah
a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)
b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)
\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)
Đúng 0
Bình luận (1)
Tính nhanh:
A = 1/3 - 3/4 - ( - 3/5 ) + 1/72 - 2/9 - 1/36 + 1/15
B = 1/ 5 - 3/7 + 5/9 - 2/11 + 7/13 - 9/16 - 7/13 + 2/11 - 5/9 + 3/7 - 1/5
C = 1/100 - 1/100 . 99 - 1/99 . 98 - 1/98 . 97 - ... - 1/3 . 2 - 1/ 2 . 1
C=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{1}{100}-\left(\frac{1}{2.1}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
=\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
=\(\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
=\(\frac{1}{100}-\frac{99}{100}\)
=\(\frac{-98}{100}=\frac{-49}{50}\)
Đúng 1
Bình luận (0)
C=1/100 -1/100.99 -1/99.98 -1/98.97-......- 1/3.2 -1/2.1
= 1/100 - (1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1)
Đặt A = 1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1 => C = 1/100 - A
Dễ thấy 1/2.1 = 1/1 - 1/2
1/3.2 = 1/2 - 1/3
.....................
1/99.98 = 1/98 - 1/99
1/100.99 = 1/99 - 1/100
=> cộng từng vế với vế ta
Đúng 0
Bình luận (0)
\(B=\left(\frac{3}{7}+\frac{-3}{7}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{5}{9}+\frac{-5}{9}\right)+\left(\frac{2}{11}-\frac{2}{11}\right)\)
\(+\left(\frac{7}{13}-\frac{7}{13}\right)-\frac{9}{16}\)
\(=0+0+0+0-\frac{1}{16}\)
\(=\frac{-1}{16}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời