51.49+52.48+...+99.1
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11 tháng 9 2021
a: \(68^2+64\cdot68+32^2\)
\(=\left(68+32\right)^2\)
=10000
b: \(51\cdot49=50^2-1=2499\)
NB
3
KT
11 tháng 8 2018
a) mk chỉnh đề
\(978^2+22^2+978.44=\left(978+22\right)^2=100^2=10000\)
b) \(52.48=\left(50+2\right)\left(50-2\right)=50^2-2^2=2496\)
c) mk chỉnh đề
\(821^2+321^2-821.642=\left(821-321\right)^2=500^2=250000\)
11 tháng 8 2018
a) \(978^2+22^2+978.44=978^2+2.22.978+22^2=\left(978+22\right)^2=1000^2=1000000\)
b) \(52.48=\left(50+2\right).\left(50-2\right)=50^2-2^2=2500-4=2496\)
c) Có nhầm dấu không bạn ơi. Mình sửa luôn nha
8212 + 3212 - 821.642 = 8212 - 2.821.321 + 3212 = (821-321)2 = 5002 = 250 000
17 tháng 5 2015
1.99+2.98+3.97+...+98.2+99.1=1.99+2.(99-1)+3.(99-2)+...+98.(99-97)+99.(99-98)
=1.99+2.99-1.2+3.99-2.3+...+98.99-97.98+99.99-98.99
=(1.99+2.99+3.99+...+98.99+99.99)-(1.2+2.3+3.4+...+98.99)
=99.(1+2+...+99)-(1.2+2.3+...+98.99)=99.4950-(1.2+2.3+...+98.99)=490050-(1.2+2.3+...+98.99)
đặt A=1.2+2.3+...+98.99
=>3A=1.2.3+2.3.3+...+98.99.3
=1.2.3+2.3.(4-1)+...+98.99.(100-97)
=1.2.3-1.2.3+2.3.4-2.3.4+...+97.98.99-97.98.99+98.99.100=98.99.100
=>A=98.99.100:3=323400
=>1.99+2.98+3.97+...+98.2+99.1=490050-323400=166650
MT
17 tháng 5 2015
1.99+2.98+3.97+4.96+...+98.2+99.1
=1.99+2.(99-1)+3.(99-2)+...+98.(99-97)+99.(99-98)
=1.99+2.99-1.2+3.99-2.3+...+98.99-97.98+99.99-98.99
=(1.99+2.99+3.99+4.99+...+98.99+99.99)-(1.2+2.3+3.4+...+97.98+98.99)
=(1+2+3+4+...+98+99).99-(98.99.100)/3
={(99-1+1)/2}.100.99-(98.99.100)/3
=49,5.100.99-(98.99.100)/3
=4950.99-(98.99.100)/3
=4950.3.33-98.100.33
B=14850.33-9800.33
B=(14850-9800).33
B=5050.33
B=166650
24 tháng 4 2015
ok !!!
giải :
ta có \(\frac{1}{15}\times\frac{1}{35}\times\frac{1}{63}\times\frac{1}{99}\times\frac{1}{143}\)
=\(\frac{1}{3\times5}\times\frac{1}{5\times7}\times\frac{1}{5\times7}\times\frac{1}{7\times9}\times\frac{1}{9\times11}\times\frac{1}{11\times13}\)
=2(\(\frac{1}{3\times5}\times\frac{1}{5\times7}\times\frac{1}{5\times7}\times\frac{1}{7\times9}\times\frac{1}{9\times11}\times\frac{1}{11\times13}\))
=\(\frac{2}{3\times5}\times\frac{2}{5\times7}\times\frac{2}{7\times9}\times\frac{2}{9\times11}\times\frac{2}{11\times13}\)
=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\)
=\(\frac{1}{3}-\frac{1}{13}\)
=\(\frac{10}{39}\)
**** cho mình nha nguyen truong giang ^^mình làm đúng đấy !!!!!!!
24 tháng 4 2015
\(=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}\)
\(=2\cdot\frac{1}{2}\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\cdot\left[\left(\frac{1}{3}-\frac{1}{13}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{11}-\frac{1}{11}\right)\right]\)
\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}\cdot\left(\frac{13}{39}-\frac{3}{39}\right)=\frac{1}{2}\cdot\frac{10}{39}=1\cdot\frac{5}{39}=\frac{5}{39}\)
BN
14 tháng 6 2016
Ta có :
C = 1.99 + 2.(99 - 1) + 3.(99 - 2) + ... + 98.(99 - 97) + 99.(99 - 98)
C = 99.(1 + 2 + 3 + ... + 98 + 99) - (2 + 2.3 + 3.4 + ...+97.98 + 98.99)
C = 99.(1 + 99).99/2 - 98.99.100/3
C = 99.50.99 - 98.33.100
C = 490050 - 323400
C = 166650
QD
14 tháng 6 2016
1.99+2.98+3.97+...+98.2+99.1=1.99+2.(99-1)+3.(99-2)+...+98.(99-97)+99.(99-98)
=1.99+2.99-1.2+3.99-2.3+...+98.99-97.98+99.99-98.99
=(1.99+2.99+3.99+...+98.99+99.99)-(1.2+2.3+3.4+...+98.99)
=99.(1+2+...+99)-(1.2+2.3+...+98.99)=99.4950-(1.2+2.3+...+98.99)=490050-(1.2+2.3+...+98.99)
đặt A=1.2+2.3+...+98.99
=>3A=1.2.3+2.3.3+...+98.99.3
=1.2.3+2.3.(4-1)+...+98.99.(100-97)
=1.2.3-1.2.3+2.3.4-2.3.4+...+97.98.99-97.98.99+98.99.100=98.99.100
=>A=98.99.100:3=323400
=>1.99+2.98+3.97+...+98.2+99.1=490050-323400=166650
6 tháng 8 2016
C=1.99+2.98+3.97+98.2+99.1
C=(1x1x99x99)+(2x98x98x2)+3x97
C=9801+9604+291
C=19405+291
C=19696