ta có : 2+4+6+...+[2xX]=2550
2 + 4 + 6 + ... + 2x = 2550
=> 2(1 + 2 + 3 + ... + x) = 2550
=> 1 + 2 + 3 + ... + x = 1275
=> x(x + 1) : 2 = 1275
=> x(x + 1) = 2550
=> x(x + 1) = 50.51
=> x = 50
ta có : 2+4+6+...+[2xX]=2550
2 + 4 + 6 + ... + 2x = 2550
=> 2(1 + 2 + 3 + ... + x) = 2550
=> 1 + 2 + 3 + ... + x = 1275
=> x(x + 1) : 2 = 1275
=> x(x + 1) = 2550
=> x(x + 1) = 50.51
=> x = 50
[ 2 CONG X ] CONG [ 4 CONG X ] CONG [ 6 CONG X ] CONG .......................... CONG [ 52 CONG X ] = 780
A] X-1,25x4= 2,5 b] [ X CONG 1 ] CONG [ X CONG 2] CONG [ X CONG 3 CONG ] .............. CONG [ X CONG 20] = 750 C ] [ X CONG 1] CONG [X CONG 2 ] CONG [ X CONG 3 ] CONG ........... CONG [ X CONG 10] = 175
[ X 1 cong ] cong [ X 2 cong ] cong[ X3 cong] CONG .............. CONG 20 = 750
x cong 1 cong x cong 2 cong ...x cong 100 bang 5050
1 CONG 3 CONG 5 CONG .........CONG [ 2 x X CONG 1] = 169
C] [ X-1/2 ] CHIA [ 1/2 CONG 1/6 CONG 1/12 CONG 1/20 ] = 1/3
1 CONG 2 CONG 3 CONG ........... CONG X = 210
1) (-1005).(x cong 2 ) bang 0
2) x cong x cong x cong 91 bang -2
3) / 5x cong 1/ bang 11
( thong cam nha , may mik k viet dc dau cong vs bang )
mik dg gap , giup mik nhe !
chứng minh ý a 8mu10 cong 2 mũ 20 chia hết cho41
chứng minh ý b 31 mu36 nhận 36 -313 mu 5 nhận 299 chia hết ch
chung minhý c 3 mu n công 3 cộng 2 mu n công 3 công 2 mu n công 2 công 2 mu n công 2 chia hết 6
chung minh y d d=2 cong 2 mu 2 cong 2mu 3 cong 2 mu 4 cong 2 mu 5 cong 2 mu 6 cong ....cong 2 mu 58 cong 2 mu 59 cong 2 mu 60 chia het cho 31
chung minh y e, e= 1cong 3 cong 3 mu 2 cong 3 mu 3 cong..cong 3 mu 98 cong 3 mu 99 chung minh e chia het cho5 chia het cho11