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a) tự làm
b) (57-17)+(58-18)+(59-19)+(60-20)+(61-21)=40+40+40+40+40=200
c) (9+15)+(11+13)+(-10-14)+(-12-12-4)=24+24-24-24-4=-4
d)-(1+2+3+.....+2007)=\(-\dfrac{\left(1+2007\right).2007}{2}=-2015028\)
a) (371-271)+(731-531)=100+200=300
b) (57-17)+(58-18)+(59-19)+(60-20)+(61-21)=40+40+40+40+40=200
c) (9+15)+(11+13)+(-10-14)+(-12-12-4)=24+24-24-24-4=-4
d)-(1+2+3+.....+2007)=−(1+2007).20072=−2015028
Tính D, biết
D=\(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+........-\frac{1}{2^{58}}\)
\(D=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+...+\frac{1}{2^{55}}-\frac{1}{2^{58}}\)
\(\Rightarrow2^3D=2^2-\frac{1}{2}+\frac{1}{2^4}-\frac{1}{2^7}+....+\frac{1}{2^{52}}-\frac{1}{2^{55}}\)
\(\Rightarrow8D+D=2^2-\frac{1}{2^{58}}\)
\(\Rightarrow D=\frac{2^2-\frac{1}{2^{58}}}{9}\)
1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + 12 + ... + 61 - 62 - 63 + 64 ( 64 số )
= ( 1 - 2 - 3 + 4 ) + ( 5 - 6 - 7 + 8 ) + ( 9 - 10 - 11 + 12 ) + ... + ( 61 - 62 - 63 + 64 ) ( 16 nhóm )
= 0 + 0 + 0 + ... + 0 ( 16 số 0 )
= 0 . 16
= 0
\(\frac{1}{2^3}\)D= \(\frac{1}{2^4}-\frac{1}{2^7}+\frac{1}{2^{10}}-\frac{1}{2^{13}}+...+\frac{1}{2^{58}}-\frac{1}{2^{61}}\)
D+ \(\frac{1}{2^3}\)D=\(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^7}-\frac{1}{2^{10}}+\frac{1}{2^{10}}+...-\frac{1}{2^{58}}+\frac{1}{2^{58}}-\frac{1}{2^{61}}\)
\(\frac{9}{8}\)D= \(\frac{1}{2}-\frac{1}{2^{61}}\)=> D= \(\frac{\frac{1}{2}-\frac{1}{2^{61}}}{\frac{9}{8}}\)
Ta có :
\(\frac{1}{13}< \frac{1}{12};\frac{1}{14}< \frac{1}{12};\frac{1}{15}< \frac{1}{12}\Rightarrow\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{1}{4}\)
\(\frac{1}{61}< \frac{1}{60};\frac{1}{62}< \frac{1}{60};\frac{1}{63}< \frac{1}{60}\Rightarrow\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=\frac{1}{20}\)
\(\Rightarrow D=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)
Vậy \(D< \frac{1}{2}\)
\(D=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)\)
Nhận xét: \(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
\(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=\frac{3}{60}=\frac{1}{20}\)
\(\Rightarrow D< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)
Vậy D < 1/2
a) $371+731-271-531$
$=(371-271)+(731-531)$
$=100+200$
$=300$
b) $57+58+59+60+61-17-18-19-20-21$
$=(57-17)+(58-18)+(59-19)+(60-20)+(61-21)$
$=40+40+40+40+40$
$=40\cdot5$
$=200$
c) $9-10+11-12+13-14+15-16$
$=(9-10)+(11-12)+(13-14)+(15-16)$
$=-1+(-1)+(-1)+(-1)$
$=-1\cdot4$
$=-4$
$\text{#}Toru$