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nhiều quá cậu ơi
mk ko muốn làm
nhìn là thấy chán rồi
ai cùng chung quan điểm với mk ko???
Ta có A= -1+7+(-72)+73+(-74)+....+72008 +72008
A.7=[-7+72+(-73)+74+....+72009 +72009] + [ -1+7+(-72)+73+(-74)+....+72008 +72008]
A.7=[72009.2+(-1) +72008] :7
b;c làm tương tự
A = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... + 2005 + 2006 - 2007 - 2008 + 2009 + 2010 ( có 2010 số )
A = ( 1 + 2 - 3 - 4 ) + ( 5 + 6 - 7 - 8 ) + .... + ( 2005 + 2006 - 2007 - 2008 ) + ( 2009 + 2010 )
A = ( - 4 ) + ( - 4 ) + ... + ( - 4 ) + 4019 ( có 503 số )
A = ( - 4 ) . 502 + 4019
A = - 2008 + 4019
A = 2011.
CHÚC LÀM BÀI VUI VẺ
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(A=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(A=7\cdot\frac{3}{35}=\frac{21}{35}\)
\(A=\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+\frac{7}{12\cdot13}+...+\frac{7}{69\cdot70}\)
\(A=7\left(\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}+...+\frac{1}{69\cdot70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\cdot\frac{3}{35}=\frac{3}{5}\)
\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)
\(B=\frac{1}{2}\left(\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)
\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{2008\cdot2010}\)
\(C=\frac{4}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)
\(C=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=2\left(\frac{1}{2}-\frac{1}{2010}\right)=2\cdot\frac{502}{1005}=\frac{1004}{1005}\)
Bài làm:
\(A=1-2+3-4+5-...-2008+2009\)
\(A=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(2007-2008\right)+2009\)
\(A=-1-1-1-...-1+2009\)(1004 số -1)
\(A=-1004+2009=1005\)
\(B=1+2-3-4+5+6-7-...-2007-2008+2009+2010\)
\(B=1+\left(2-3-4+5\right)+\left(6-7-8+9\right)+...+\left(2006-2007-2008+2009\right)+2010\)
\(B=1+0+0+...+0+2010\)
\(B=2011\)
Học tốt!!!!
a, C=-1+3-5+7-9+...-2009+2011-2013
=(-1+3)+(-5+7)+...+(-2009+2011)-2013
=(2+2+2+...+2)-2013 (503 số 2)
=2.503-2013=1006-2013=-1007
Vậy C=-1007
b, D=2-4+6-8+...+2006-2008+2010-2012
=(2-4)+(6-8)+...+(2006-2008)+(2010-2012) (503 cặp số)
=-2+(-2)+...+(-2)+(-2) (503 số -2)
=-2.503=-1006
Vậy D=-1006
c, G=1+2-3-4+5+6-7-8+...-111-112+113+114+115
=(1+2-3-4)+(5+6-7-8)+...+(109+110-111-112)+(113+114+115)
=-4+(-4)+...+(-4)+342 (28 số -4)
=-4.28+342=-112+342=230
Vậy G=230
Mình làm bài 1, bài 2 bạn tự làm nhé!
Bài 1:
a) \(2\left(4x-8\right)-7\left(3+x\right)=\left|4\right|\left(3-2\right)\)
\(\Leftrightarrow2\left(4x-8\right)-7\left(3+x\right)=4.1\)
\(\Leftrightarrow2\left(4x-8\right)-7\left(3+x\right)=4\)
\(\Leftrightarrow8x-16-21-7x=4\)
\(\Leftrightarrow x-37=4\)
\(\Leftrightarrow x=4+37\)
\(\Leftrightarrow x=41\)
Vậy \(x=41\)
b) \(8\left(x-\left|-7\right|\right)-6\left(x-2\right)=\left|-8\right|.6-50\)
\(\Leftrightarrow8\left(x-7\right)-6\left(x-2\right)=\left|-8\right|.6-50\)
\(\Leftrightarrow8\left(x-7\right)-6\left(x-2\right)=8.6-50\)
\(\Leftrightarrow8\left(x-7\right)-6\left(x-2\right)=48-50\)
\(\Leftrightarrow8\left(x-7\right)-6\left(x-2\right)=-2\)
\(\Leftrightarrow8x-56-6x+12=-2\)
\(\Leftrightarrow2x-44=-2\)
\(\Leftrightarrow2x=-2+44\)
\(\Leftrightarrow2x=42\)
\(\Leftrightarrow x=42:2\)
\(\Leftrightarrow x=21\)
Vậy \(x=21\)
