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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
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
Ta có :
\(A=\frac{1}{3}-\frac{3}{4}-\left(-\frac{3}{5}\right)+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow A=\frac{5}{15}-\frac{54}{72}+\frac{9}{15}+\frac{1}{72}-\frac{16}{72}-\frac{1}{72}+\frac{1}{15}\)
\(\Rightarrow A=\left(\frac{5}{15}+\frac{9}{15}+\frac{1}{15}\right)+\left(-\frac{54}{72}+\frac{1}{72}-\frac{16}{72}-\frac{2}{72}\right)\)
\(\Rightarrow A=1-\frac{71}{72}=\frac{1}{72}\)
\(\dfrac{3}{7}\cdot\dfrac{9}{26}-\dfrac{1}{14}\cdot\dfrac{1}{13}\)
\(=\dfrac{3\cdot9-1}{14\cdot13}=\dfrac{26}{14\cdot13}=\dfrac{2}{14}=\dfrac{1}{7}\)
Ta có :
\(\frac{3}{8}.\frac{58}{3}-\frac{3}{8}.\frac{100}{3}\)
\(=\frac{3}{8}.\left(\frac{58}{3}-\frac{100}{3}\right)\)
\(=\frac{3}{8}.\frac{-42}{3}\)
\(=\frac{-42}{8}=\frac{-21}{4}\)
\(\frac{4^{19}+8^7}{256^4+32^2}=\frac{\left(2^2\right)^{19}+\left(2^3\right)^7}{\left(2^8\right)^4+\left(2^5\right)^2}=\frac{2^{38}+2^{21}}{2^{32}+2^{10}}=\frac{2^{21}.\left(2^{17}+1\right)}{2^{10}.\left(2^{22}+1\right)}=\frac{2^{11}.\left(2^{17}+1\right)}{2^{22}+1}=\frac{2^{28}+2^{11}}{2^{22}+1}\)
\(\dfrac{5}{3}+\left(\dfrac{-2}{7}\right)-\left(\dfrac{-1}{2}\right)\)
= \(\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{1}{2}\)
= \(\dfrac{70}{42}-\dfrac{12}{42}+\dfrac{21}{42}\)
= \(\dfrac{79}{42}\)
\(\dfrac{5}{3}+\dfrac{-2}{7}-\dfrac{-1}{2}=\dfrac{98}{42}-\dfrac{12}{42}+\dfrac{21}{42}=\dfrac{107}{42}\)