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26 tháng 7 2018
\(\frac{11}{9}+\frac{35}{18}\)
\(=\frac{19}{6}\)
\(\frac{22}{21}:\frac{1}{2}+\frac{25}{12}:\frac{1}{2}\)
\(\frac{44}{21}+\frac{25}{6}\)
\(\frac{263}{42}\)
8 tháng 6 2023
\(\dfrac{1}{3\times7}+\dfrac{1}{7\times11}+\dfrac{1}{11\times15}+...+\dfrac{1}{a\times\left(a+4\right)}=\dfrac{50}{609}\)
\(\dfrac{1}{4}\times\left(\dfrac{4}{3\times7}+\dfrac{4}{7\times11}+...+\dfrac{4}{a\times\left(a+4\right)}\right)=\dfrac{50}{609}\)
\(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{a}-\dfrac{1}{a\times4}=\dfrac{50}{609}\div\dfrac{1}{4}\)
\(\dfrac{1}{3}-\dfrac{1}{a\times4}=\dfrac{200}{609}\)
\(\dfrac{1}{a\times4}=\dfrac{1}{3}-\dfrac{200}{609}\)
\(\dfrac{1}{a\times4}=\dfrac{1}{203}\)
\(a\times4=203\)
\(a=\dfrac{203}{4}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 6 2023
\(\dfrac{1}{3\times7}\)+\(\dfrac{1}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{1}{a\times\left(a+4\right)}\) = \(\dfrac{50}{609}\)
4\(\times\)( \(\dfrac{1}{3\times7}\) +\(\dfrac{1}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{1}{a\times\left(a+4\right)}\)) = \(\dfrac{50}{609}\) \(\times\)4
\(\dfrac{4}{3\times7}\)+ \(\dfrac{4}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{4}{a\times\left(a+4\right)}\) = \(\dfrac{50}{609}\) \(\times\) 4
\(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\)-\(\dfrac{1}{15}\)+...+\(\dfrac{1}{a}\)-\(\dfrac{1}{a+4}\) = \(\dfrac{200}{609}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{a+4}\) = \(\dfrac{200}{609}\)
\(\dfrac{1}{a+4}\) = \(\dfrac{1}{3}\) - \(\dfrac{200}{609}\)
\(\dfrac{1}{a+4}\) = \(\dfrac{1}{203}\)
a + 4 = 203
\(a\) = 203 - 4
\(a\) = 199
Đáp số: \(a\) = 199
13 tháng 3 2016
\(\left(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}\right)x\frac{9}{15}:\frac{7}{12}=\frac{1}{4}x\frac{9}{15}:\frac{7}{12}=\frac{3}{20}:\frac{7}{12}=\frac{9}{35}\)
ai hminh tích lại nhà
N
25 tháng 7 2018
a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 )
vì ( 125125 x 127 - 127127 x 125 ) =[125125 x (125+2)] - 127127 x 125 ) =>125125 x (125+2)=125.125125+125125.2=125125.125+250250=125125.125+125.2002=125.(125125+2002)=125.127127
=> ( 125125 x 127 - 127127 x 125 )=127127.125-127127.125=0
=> (1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) =0
25 tháng 7 2018
a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 )
= ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x 0
= 0
b, \(\frac{1}{3}\)+ \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ \(\frac{1}{143}\)+ \(\frac{1}{195}\)
= \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+\(\frac{1}{7}\)- \(\frac{1}{9}\)+...........+\(\frac{1}{13}\)- \(\frac{1}{15}\)
= \(\frac{1}{3}\)- \(\frac{1}{15}\)
= \(\frac{4}{15}\)
19 tháng 6 2015
A= 7/5*7 + 7/7*9 + ... + 7/53*55
A= 7/2*( 2/5*7 + 2/7*9 + ... + 2/53*55 )
A= 7/2*( 7-5/5*7 + 9-7/7*9 + ... + 55-53/53*55 )
A= 7/2*( 1/5-1/7 + 1/7-1/9 + ... + 1/53-1/55 )
A= 7/2*( 1/5-1/55 )
A= 7/2*2/11
A= 7/11
A= 7/11 > 1/2
Nên: A > 1/2
B= 1/3 + 1/15 + 1/35 + ... + 1/99
B= 1/1*3 + 1/3*5 + 1/5*7 + ... + 1/9*11
B= 2*( 2/1*3 + 2/3*5 + 1/5*7 + ... + 2/9*11 )
B= 2*( 3-1/1*3 + 5-3/3*5 + 7-5/5*7 + ... + 11-9/9*11 )
B= 2*( 1/1-1/3 + 1/3-1/5 + 1/5-1/7 + ... + 1/9-1/11 )
B= 2*( 1/1-1/11 )
B= 2*10/11
B= 20/11
B= 20/11 < 1/2
Nên: B < 1/2
20 tháng 6 2015
A= 7/5*7 + 7/7*9 + ... + 7/53*55
A= 7/2*( 2/5*7 + 2/7*9 + ... + 2/53*55 )
A= 7/2*( 7-5/5*7 + 9-7/7*9 + ... + 55-53/53*55 )
A= 7/2*( 1/5-1/7 + 1/7-1/9 + ... + 1/53-1/55 )
A= 7/2*( 1/5-1/55 )
A= 7/2*2/11
A= 7/11
A= 7/11 > 1/2
Nên: A > 1/2
B= 1/3 + 1/15 + 1/35 + ... + 1/99
B= 1/1*3 + 1/3*5 + 1/5*7 + ... + 1/9*11
B= 2*( 2/1*3 + 2/3*5 + 1/5*7 + ... + 2/9*11 )
B= 2*( 3-1/1*3 + 5-3/3*5 + 7-5/5*7 + ... + 11-9/9*11 )
B= 2*( 1/1-1/3 + 1/3-1/5 + 1/5-1/7 + ... + 1/9-1/11 )
B= 2*( 1/1-1/11 )
B= 2*10/11
B= 20/11
B= 20/11 < 1/2
Nên: B < 1/2
Máy tính
\(1\frac{1}{3}:1\frac{1}{7}:1\frac{1}{15}\)
\(=\frac{4}{3}:\frac{8}{7}:\frac{16}{15}\)
\(=\frac{7}{6}:\frac{16}{15}\)
\(=\frac{35}{32}\)