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\(\dfrac{3}{7}\cdot\dfrac{9}{26}-\dfrac{1}{14}\cdot\dfrac{1}{13}\)
\(=\dfrac{3\cdot9-1}{14\cdot13}=\dfrac{26}{14\cdot13}=\dfrac{2}{14}=\dfrac{1}{7}\)
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
a)(-7 / 5 + 3 / 8 ) : 2009 / 2010 + (-3 / 5+5 / 8) : 2009 / 2010
=[ (-7 / 5 + 3 / 8 ) + (-3 / 5+5 / 8) ] : 2009/2010
=[ -7/5 +3/8 + (-3)/5+5/8 ] : 2009/2010
=[ (-7/5 + (-3)/5) + (3/8 + 5/8) ] :2009/2010
=[-2+1] : 2009/2010
=-1 :2009/2010
=-2009/2010
b)\(\frac{9^8\cdot4^3}{27^4\cdot6^5}=\frac{\left(3^2\right)^8\cdot\left(2^2\right)^3}{\left(3^3\right)^4\cdot\left(2\cdot3\right)^5}=\frac{3^{16}\cdot2^6}{3^{12}\cdot2^5\cdot3^5}=\frac{3^{16}\cdot2^5\cdot2}{3^{16}\cdot3^1\cdot2^5}=\frac{2}{3^1}=\frac{2}{3}\)
\(\frac{6}{5}-\frac{1}{4}+\frac{4}{5}-\frac{3}{4}\)
\(=\left(\frac{6}{5}+\frac{4}{5}\right)-\left(\frac{1}{4}+\frac{3}{4}\right)\)
\(=2-1\)
\(=1\)
\(\frac{7}{9}-\frac{5}{12}+\frac{2}{9}-\frac{7}{12}\)
\(=\left(\frac{7}{9}+\frac{2}{9}\right)-\left(\frac{5}{12}+\frac{7}{12}\right)\)
\(=1-1\)
\(=0\)
các câu sau tương tự
A=1
B=0
C=ÂM 8/15
D=0
E=2
F=3/2
H=2/3
LẦN SAU CHO KHÓ SÍU NHA LỚP 5 CŨNG LÀM ĐC
Ta có :
\(A=\frac{1}{3}-\frac{3}{4}-\left(-\frac{3}{5}\right)+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow A=\frac{5}{15}-\frac{54}{72}+\frac{9}{15}+\frac{1}{72}-\frac{16}{72}-\frac{1}{72}+\frac{1}{15}\)
\(\Rightarrow A=\left(\frac{5}{15}+\frac{9}{15}+\frac{1}{15}\right)+\left(-\frac{54}{72}+\frac{1}{72}-\frac{16}{72}-\frac{2}{72}\right)\)
\(\Rightarrow A=1-\frac{71}{72}=\frac{1}{72}\)
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 :/