Tính
20^2-19^2+18^2-17^2+...+2^2-1^2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{\dfrac{19}{1}+\dfrac{18}{2}+\dfrac{17}{3}+....+\dfrac{1}{19}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{1+\left(\dfrac{18}{2}+1\right)+\left(\dfrac{17}{3}+1\right)+\left(\dfrac{1}{19}+1\right)}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{1+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{20}{19}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{20.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}\)
\(=\dfrac{1}{20}\)
* Cách làm : Tử giữ nguyên,còn mẫu ta biến đổi như sau:
Mẫu : ( \(\frac{19}{1}\)+ 1 ) + ( \(\frac{18}{2}\)+ 1 ) + ( \(\frac{17}{3}\)+ 1 ) +...+ ( \(\frac{3}{17}\)+ 1 ) + ( \(\frac{2}{18}\)+ 1 ) + ( \(\frac{1}{19}\)+ 1 ) - 19 ( vì ta cộng với 19 số 1 nên phải trừ 19 )
= \(\frac{20}{1}\)+ \(\frac{20}{2}\)+ \(\frac{20}{3}\)+...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)- 19
= \(\frac{20}{2}\)+ \(\frac{20}{3}\)+...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)+ ( \(\frac{20}{1}\)- 19)
= \(\frac{20}{2}\)+ \(\frac{20}{3}\)+ ...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)+ \(\frac{20}{20}\)
= 20.( \(\frac{1}{2}\)+ \(\frac{1}{3}\)+...+ \(\frac{1}{17}\)+ \(\frac{1}{18}\)+ \(\frac{1}{19}\)+ \(\frac{1}{20}\))
=> \(\frac{Tử}{Mâu}\)= \(\frac{1}{20}\)
Phùng Quang Thịnh biến đổi sai 1 chỗ kìa
-19 = \(\frac{20}{20}-20\)chứ mà bạn
B = 2^20 - ( 2^19 + 2^18 + 2^17 + ... + 2^1 + 2^0
B = (2^19 x 2 - 2^19) - 2^18 - 2^17 - ... - 2^1 - 0
B = (2^18 x 2 - 2^18) - 2^17 - 2^16 - ... - 2^1 - 0
B = (2^17 x 2 - 2^17) - 2^16 - 2^15 - ... - 2^1 - 0
Cứ tách xong lại đóng ngoặc cho đến:
B = 2^1 x 2 - 2^1 - 0
B = 2^1 - 0
B = 2
Đặt A=219+218+...+20
2A=2(219+218+...+1)
2A=220+219+...+2
2A-A=(220+219+...+2)-(219+218+...+2)
A=220-2
Thay A vào B ta có: 220-(220-2)
=220-220+2
=0+2=2
Ta có: 2B = \(2^{21}-2^{20}-...-2^0\)
B = \(2^{20}-2^{19}-...-2^1\)
=> 2B + B = \(2^{21}-1\)
=> 3B = \(2^{21}-1\)
=> B = \(\frac{2^{21}-1}{3}\)
2^20 - 2^19 - 2^18 - 2^17 - ... - 2^1 - 2^0
=(2^19 x 2 - 2^19) - 2^18 - 2^17 - 2^16 - ... - 2^1
=(2^18 x 2 - 2^18) - 2^17 - 2^16 - 2^15 - ... - 2^1
Cứ tính xong trong ngoặc lại ghép tiếp........
=2^1 x 2 - 2^1
=2^1
=2
Ta có phần tử \(=\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+...+\frac{18}{2}+\frac{19}{1}\)
\(=\left(\frac{1}{19}+1\right)+\left(\frac{2}{18}+1\right)+...+\left(\frac{18}{2}+1\right)+\left(\frac{19}{1}+1\right)-19\)
\(=\frac{20}{19}+\frac{20}{18}+...+\frac{20}{2}+\frac{20}{1}+\frac{20}{20}-20\)
\(=20.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{19}+\frac{1}{20}\right)\left(1\right)\)
Thay (1) vào P ta được :
\(P=\frac{20.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}}\)
\(=20\)
\(\left(20^2+18^2+16^2+......+4^2+2^2\right)-\left(19^2+17^2+.....+3^2+1^2\right)\)
\(=20^2-19^2+18^2-17^2+......+2^2-1^2\)
\(=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+.......+\left(2-1\right)\left(2+1\right)\)
\(=39+35+....+7+3\)
\(=\left(39+3\right)\left[\left(39-3\right):4+1\right]:2=210\)
\(20^2-19^2+18^2-17^2+...+2^2-1^2\)
\(=\left(20-19\right)\left(20+19\right)+\left(18-17\right)\left(18+17\right)+...\left(2-1\right)\left(2+1\right)\)
\(=39+35...+3\)
Đến đây ok r nhá