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\(\frac{-1}{91}+\frac{-1}{247}+\frac{-1}{475}+\frac{-1}{775}+\frac{-1}{1147}\)
\(=-\left(\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\right)\)
\(=-[\frac{1}{6}.\left(\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{31.37}\right)]\)
\(=-\text{[}\frac{1}{6}.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\text{]}\)
\(=-\text{[}\frac{1}{6}.\left(\frac{1}{7}-\frac{1}{37}\right)\text{]}\)
\(=-\text{[}\frac{1}{6}.\frac{30}{259}\text{]}\)
\(=-\frac{5}{259}\)
\(\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)
\(=\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{37}\right)\)
\(=\dfrac{1}{6}\left(1-\dfrac{1}{37}\right)\)
\(=\dfrac{1}{6}.\dfrac{36}{37}\)
\(=\dfrac{6}{37}\)
\(#Wendy.Dang\)
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\)
Dễ thấy: \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\Rightarrow x+2004=0\Leftrightarrow x=-2014\)
Ta có :
\(A=\frac{1}{2003\cdot2002}-\frac{1}{2002\cdot2001}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
\(A=-\left(\frac{1}{2003\cdot2002}+\frac{1}{2002\cdot2001}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)
\(A=-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2001\cdot2002}+\frac{1}{2002\cdot2003}\right)\)
\(A=-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2003}\right)\)
\(A=-\left(1-\frac{1}{2003}\right)\)
\(A=-\frac{2002}{2003}\)
\(A=\frac{1}{2003.2002}-\frac{1}{2002.2001}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2001.2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2001}-\frac{1}{2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)
\(=-\left(1-\frac{1}{2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)
\(=-1+\frac{1}{2002}.+\frac{1}{2002}.\frac{1}{2003}\)
\(=-1+\frac{1}{2002}\left(1+\frac{1}{2003}\right)\)
\(=-1+\frac{1}{2002}.\frac{2004}{2003}\)
\(=-1+\frac{2}{2003}\)
\(=\frac{-2003+2}{2003}\)
\(=\frac{-2001}{2003}\)
bài 1)
70:2=35(m)
Gọi a và b lần lượt là chiều rộng và chiều dài của miếng đất
Từ b/a = 4 /3 = > 3/a = 4 /b
= > 3/ a = 4/ b = 3 + 4/ a + b = 7/ 35 = 5 /3 a = 5
= > a = 3.5 = 15/ 4 b = 5
= > b = 5.4 = 20
Vậy diện tích miếng đất đó là:
15.20=300(m2)
2) Bài 138 (Sách bài tập - tập 1 - trang 33)
bài 2 cậu vào cái ý là có
\(5\frac{4}{7}\): [ x : 1,3 + 8,4 . \(\frac{6}{7}\). ( 6 - \(\frac{\left(2,3+5\div6,25\right)\times7}{8\times0,-125+6,9}\)) ] = \(1\frac{1}{14}\)
\(\frac{39}{7}\): [ x : 1,3 + \(\frac{36}{5}\). ( 6 - \(\frac{\left(2,3+0,8\right).7}{0,1+6,9}\)) ] = \(\frac{15}{14}\)
\(\frac{39}{7}\): [ x : 1,3 + \(\frac{36}{5}\). ( 6 - \(\frac{3,1.7}{7}\)) ] = \(\frac{15}{14}\)
\(\frac{39}{7}\): [ x : 1,3 + \(\frac{36}{5}\). ( 6 - 3,1 ) ] = \(\frac{15}{14}\)
x : 1,3 + \(\frac{36}{5}\). 2,9 = \(\frac{39}{7}\): \(\frac{15}{14}\)
x : 1,3 + 20,88 = 5,2
x : 1,3 = - 15,68
x = - 15,68 . 1,3
x = - 20,384
ta có
\(5\frac{4}{7}:\left\{x:1,3+8,4.\frac{6}{7}.\left[6-\frac{\left(2,3+5:6,25\right).7}{8.0,0125+6,9}\right]\right\}=1\frac{1}{14}\)
\(\Leftrightarrow\frac{39}{7}:\left\{x:1,3+7,2.\left[6-\frac{\left(2,3+0,8\right).7}{0,1+6,9}\right]\right\}=\frac{15}{14}\)
\(\Leftrightarrow\frac{39}{7}:\left\{x:1,3+7,2.\left[6-\frac{3,1.7}{7}\right]\right\}=\frac{15}{14}\)
\(\Leftrightarrow\frac{39}{7}:\left\{x:1,3+7,2.2,9\right\}=\frac{15}{14}\Leftrightarrow\left\{x:1,3+7,2.2,9\right\}=\frac{39}{7}:\frac{15}{14}\)
\(\Leftrightarrow x:1,3+20,88=5,2\Leftrightarrow x:1,3=-15,68\Leftrightarrow x=-20,384\)
\(B=\frac{1}{5}-\frac{3}{7}+\frac{5}{9}-\frac{2}{11}+\frac{7}{13}-\frac{9}{16}-\frac{7}{13}+\frac{2}{12}-\frac{5}{9}+\frac{3}{7}-\frac{1}{5}-\frac{1}{5}\)
\(B=\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{3}{7}-\frac{3}{7}\right)+\left(\frac{5}{9}-\frac{5}{9}\right)+\left(\frac{7}{13}-\frac{7}{13}\right)-\frac{2}{11}+\frac{2}{12}-\frac{9}{16}-\frac{1}{5}\)
\(B=0-0+0+0-\frac{2}{11}+\frac{2}{12}-\frac{9}{16}-\frac{1}{5}\)
\(B=\frac{-2}{11}+\frac{2}{12}-\frac{9}{16}-\frac{1}{5}\)
Đến đây chỉ còn cách quy đồng thôi
5/9 : (1/11 - 5/12) + 5/9 : (1/15 - 2/3)
= 5/9 : (-43/132) + 5/9 : (-3/5)
= 5/9 . (-132/43) + 5/9 . (-5/3)
= 5/9 . [-132/43 + (-5/3)]
= 5/9 . 445/129
= 2225/1161.
Ai k mình mình k lại.
