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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) $9-10+11-12+13-14+15-16$
$=(9-10)+(11-12)+(13-14)+(15-16)$
$=-1+(-1)+(-1)+(-1)$
$=-1\cdot4$
$=-4$
$\text{#}Toru$
Ta có :
\(A=100\left(1+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{9899}{9900}\right)\)
\(A=100\left(1+\frac{6-1}{6}+\frac{12-1}{12}+\frac{20-1}{20}+...+\frac{9900-1}{9900}\right)\)
\(A=100\left(1+\frac{6}{6}-\frac{1}{6}+\frac{12}{12}-\frac{1}{12}+\frac{20}{20}-\frac{1}{20}+...+\frac{9900}{9900}-\frac{1}{9900}\right)\)
\(A=100\left(1+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{9900}\right)\)
\(\frac{A}{100}=1+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{9900}\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\right)\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2}-\frac{1}{100}\right)\)
Do từ \(2\) đến \(99\) có \(99-2+1=98\) số nên có \(98\) số \(1\) suy ra :
\(\frac{A}{100}=98-\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\frac{A}{100}=98-\frac{49}{100}\)
\(\frac{A}{100}=\frac{9751}{100}\)
\(A=\frac{9751}{100}.100\)
\(A=9751\)
Vậy \(A=9751\)
Chúc bạn học tốt ~
\(=\frac{7}{19}.\left(\frac{8}{11}+\frac{3}{11}\right)+\frac{12}{19}\)
\(=\frac{7}{19}.1+\frac{12}{19}\)
\(=1\)
\(\frac{7}{19}\).\(\frac{8}{11}\)+\(\frac{7}{19}\).\(\frac{3}{11}\)+\(\frac{12}{19}\)
=\(\frac{7}{19}\).1+\(\frac{12}{19}\)
=1
hok tốt
ta có :
\(64^{10}-32^{11}-16^{13}=\left(2^6\right)^{10}-\left(2^5\right)^{11}-\left(2^4\right)^{13}\)
\(=2^{60}-2^{55}-2^{52}=2^{52}\left(2^8-2^3-1\right)=2^{52}\times247\)
mà 247 chia hết cho 19 nên số ban đầu chia hết cho 19
= 165