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S = ( 1 + 2 - 3 - 4 ) + ( 5 + 6 - 7 - 8 ) + ... + ( 2001 + 2001 - 2003 - 2004 ) + ( 2005 + 2006 )
S = ( - 4 ) + ( - 4 ) + .... + ( - 4 ) + ( 2005 + 2006 )
Dãy S có : 2004 - 1 : 1 + 1 = 2004 số hạng
Dãy số S : 2004 : 4 = 501 số ( - 4 )
Dãy đó S = -4 x 501 = -2004
S = -2004 + ( 2005 + 2006 )
S = -2004 + 4011
S = 2007
Ta có :
\(A=\frac{1}{3}-\frac{3}{4}-\left(-\frac{3}{5}\right)+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow A=\frac{5}{15}-\frac{54}{72}+\frac{9}{15}+\frac{1}{72}-\frac{16}{72}-\frac{1}{72}+\frac{1}{15}\)
\(\Rightarrow A=\left(\frac{5}{15}+\frac{9}{15}+\frac{1}{15}\right)+\left(-\frac{54}{72}+\frac{1}{72}-\frac{16}{72}-\frac{2}{72}\right)\)
\(\Rightarrow A=1-\frac{71}{72}=\frac{1}{72}\)
= 93/4 . 7/5 - 53/4 : 5/7
=93/4 . 7/5 - 53/4 . 7/5
= 7/5 .( 93/4 - 53/4 )
= 7/5 . 10 ( rút gọn chéo 7/5 . 10 = 7 .2 =14)
= 14
tích cho mình nha
\(\frac{4^{19}+8^7}{256^4+32^2}=\frac{\left(2^2\right)^{19}+\left(2^3\right)^7}{\left(2^8\right)^4+\left(2^5\right)^2}=\frac{2^{38}+2^{21}}{2^{32}+2^{10}}=\frac{2^{21}.\left(2^{17}+1\right)}{2^{10}.\left(2^{22}+1\right)}=\frac{2^{11}.\left(2^{17}+1\right)}{2^{22}+1}=\frac{2^{28}+2^{11}}{2^{22}+1}\)
\(B=\dfrac{1+\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}}{4+\dfrac{4}{7}+\dfrac{4}{7^2}-\dfrac{4}{7^3}}\cdot\dfrac{858585}{313131}\cdot\left(-1\dfrac{14}{17}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{85}{31}\cdot\dfrac{-31}{17}\)
\(=\dfrac{-5}{4}\)
\(A=\dfrac{10}{7.12}+\dfrac{10}{12.17}+\dfrac{10}{17.22}+...+\dfrac{10}{502.507}\) (sửa 502+507 thành 503.507)
\(\Rightarrow A=10\left(\dfrac{1}{7.12}+\dfrac{1}{12.17}+\dfrac{1}{17.22}+...+\dfrac{1}{502.507}\right)\)
\(\Rightarrow A=10.\dfrac{1}{5}\left(\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}+...+\dfrac{1}{502}-\dfrac{1}{507}\right)\)
\(\Rightarrow A=2.\left(\dfrac{1}{7}-\dfrac{1}{507}\right)=2.\left(\dfrac{500}{3549}\right)=\dfrac{1000}{3549}\)
\(B=\dfrac{4}{8.13}+\dfrac{4}{13.18}+\dfrac{4}{18.23}+...+\dfrac{4}{253.258}\)
\(\Rightarrow B=4\left(\dfrac{1}{8.13}+\dfrac{1}{13.18}+\dfrac{1}{18.23}+...+\dfrac{1}{253.258}\right)\)
\(\Rightarrow B=4.\dfrac{1}{5}\left(\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{23}+...+\dfrac{1}{253}-\dfrac{1}{258}\right)\)
\(\Rightarrow B=\dfrac{4}{5}\left(\dfrac{1}{8}-\dfrac{1}{258}\right)=\dfrac{4}{5}\left(\dfrac{129}{1032}-\dfrac{8}{1032}\right)=\dfrac{4}{5}.\dfrac{121}{1032}=\dfrac{121}{1290}\)
Bài làm
\(\frac{\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)}{\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}\right)}\)
\(=\frac{\left(\frac{2}{2}+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\right)}{\left(\frac{2}{2}-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}\right)}\)
\(=\frac{\frac{1}{2}\left(2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}\right)}{\frac{1}{2}\left(2-1+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}\right)}\)
\(=\frac{3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}}{1+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}}\)
\(=\frac{\frac{24}{8}+\frac{4}{8}+\frac{2}{8}+\frac{1}{8}}{\frac{8}{8}+\frac{4}{8}-\frac{2}{8}+\frac{1}{8}}\)
\(=\frac{31}{8}\div\frac{11}{8}\)
\(=\frac{31}{8}\cdot\frac{8}{11}\)
\(=\frac{31}{11}\)
P/S: Trông không thuận tiện lắm :/
8/15 + 7/4 + 7/15 + 5/4
= 8/15 + 7/15 + 7/4 +5/4
= ( 8/15 + 7/15) + ( 7/4 + 5/4)
= 15/15 + 12/4
= 1 + 3
=4
=8/15+7/15+7/4+5/4
=1+3
=4