Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.
\(A=\dfrac{253\cdot75-161\cdot37+253\cdot25-161\cdot63}{100\cdot47-12\cdot3,5-5,8:0,1}\\ =\dfrac{253\cdot75+253\cdot25-\left(161\cdot37+161\cdot63\right)}{100\cdot47-\left(12\cdot3,5+5,8:0,1\right)}\\ =\dfrac{253\cdot\left(75+25\right)-161\cdot\left(37+63\right)}{100\cdot47-\left(42+58\right)}\\ =\dfrac{253\cdot100-161\cdot100}{100\cdot47-100}\\ =\dfrac{100\cdot\left(253-161\right)}{100\cdot\left(47-1\right)}\\ =\dfrac{253-161}{47-1}\\ =\dfrac{92}{46}\\ =2\)
2.
a,
Quy luật: Số thứ \(n=3^{n-1}\left(1\le n\le6\right)\)
Dãy số tương đương với:
\(3^{1-1};3^{2-1};3^{3-1};...;3^{7-1}\\ \Leftrightarrow3^0;3^1;3^2;...;3^6\)
\(3^0+3^1+3^2+...+3^6\\ =1\cdot\left(3^0+3^1+3^2+...+3^6\right)\\ =\dfrac{3-1}{2}\cdot\left(3^0+3^1+3^2+...+3^6\right)\\ =\dfrac{\left(3-1\right)\cdot\left(3^0+3^1+3^2+...+3^6\right)}{2}\\ =\dfrac{3^1-3^0+3^2-3^1+3^3-3^2+...+3^7-3^6}{2}\\ =\dfrac{3^7-1}{2}\\ =\dfrac{2187-1}{2}\\ =\dfrac{2186}{2}\\ =1093\)
b,
Quy luật: Số thứ \(n=2^{n-1}\left(1\le n\le12\right)\)
Dãy số tương đương với:
\(2^{1-1};2^{2-1};2^{3-1};...;2^{12-1}\\ \Leftrightarrow2^0;2^1;2^2;...;2^{11}\)
\(2^0+2^1+2^2+...+2^{11}\\ =1\cdot\left(2^0+2^1+2^2+...+2^{11}\right)\\ =\left(2-1\right)\cdot\left(2^0+2^1+2^2+...+2^{11}\right)\\ =2^1-2^0+2^2-2^1+2^3-2^2+...+2^{12}-2^{11}\\ =2^{12}-1\\ =4096-1\\ =4095\)
c,
Quy luật: Số thứ \(n=n^2\left(1\le n\le33\right)\)
Dãy số tương đương với:
\(1^2;2^2;3^2;...;33^2\)
\(1^2+2^2+3^2+...+33^2\\ =\dfrac{33\cdot34\cdot65}{6}\\ =12155\)
Cách chứng minh: https://diendan.hocmai.vn/threads/1-2-2-2-n-2.220374/
a)387+(-224)+(-87)=[387+(-87)]+(-224)=300+(-87)=213
b)(-75)+329+(-25)=[(-75)+(-25)]+329=-100+329=229
e)251+(-114)+(-151)+(-216)=[251+(-151)]+[(-114)+(-216)]=100+(-330)=-230
j)(-137)+77+(-253)+37-47=[(-137)+37]+77+[(-253)-47]=-100+77-300=177-300=-123
47.63 = (55 - 8)(55 + 8)
= 55² - 8²
= 3025 - 64
= 2961
--------
47.(63 - 15) - 63.(47 - 15)
= 47.63 - 47.15 - 63.47 + 63.15
= 47.(63 - 63) + 15.(63 - 47)
= 47.0 + 15.16
= 0 + 240
= 240
47.63 = (55 - 8)(55 + 8)
= 55² - 8²
= 3025 - 64
= 2961
--------
47.(63 - 15) - 63.(47 - 15)
= 47.63 - 47.15 - 63.47 + 63.15
= 47.(63 - 63) + 15.(63 - 47)
= 47.0 + 15.16
= 0 + 240
= 240
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2007}-\frac{1}{2008}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{2007}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2008}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}+\frac{1}{2008}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2008}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{1004}\)
\(A=\frac{1}{1005}+\frac{1}{1006}+\frac{1}{1007}+...+\frac{1}{2008}\) (1)
\(B=\frac{1}{1005}+\frac{1}{1006}+\frac{1}{1007}+...+\frac{1}{2008}\) (2)
\(\left(1\right)\left(2\right)\Rightarrow\frac{A}{B}=\frac{\frac{1}{1005}+\frac{1}{1006}+\frac{1}{1007}+...+\frac{1}{2008}}{\frac{1}{1005}+\frac{1}{1006}+\frac{1}{1007}+...+\frac{1}{2008}}=1\)
a) (-1) + 2 + (-3) + 4 + .... + (-2009) + 2010
= (-1 + 2) + (-3 + 4) + ..... + (-2009 + 2010)
= -1 + (-1) + (-1) + .... + (-1)
= -1 . 1005 = -1005
b) 1 + (-2) + (-3) + 4 + 5 + (-6) + (-7) + 8 + ... + 2005 + (-2006) + (-2007) + 2008
= [1 + (-2) + (-3) + 4] + [5 + (-6) + (-7) + 8 ] + ..... + [2005 + (-2006) + (-2007) + 2008]
= 0 + 0 + ...... + 0 = 0