5 mu13 : 5mu10 - 25*2mu2
5mu13 : 5mu10 - 25*2mu2
513 : 510 - 25 * 22
= 513-10 - 25 * 4
= 53 - 100
= 125 - 100
= 25
A=1+5+5mu2+5mu3...+5mu10+5mu11chia het ch 6 va31
\(S=\dfrac{2mu2}{1.2}+\dfrac{2mu2}{2.3}+\dfrac{2mu2}{3.4}+...+\dfrac{2mu2}{2022.2023}\)
(mu = mũ)
\(S=\dfrac{2^2}{1.2}+\dfrac{2^2}{2.3}+\dfrac{2^2}{3.4}+...+\dfrac{2^2}{2022.2023}\)
\(S=2^2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)
\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2023}\right)\)
\(S=2^2.\dfrac{2022}{2023}\)
\(S=\dfrac{2^2.2022}{2023}=\dfrac{8088}{2023}\)
A=3mu2+3mu2+2mu2+2mu2+.........=1mu2+1mu2
Cho S =1+2+2mu2+2mu3+...+2mu9+2mu10+2mu11.Hay so sanh Svoi 5*2mu10
Cho b=1+2+2mu2+2mu3+...+2mu6,A=2mu2+2mu3+2mu4+..+2mutam chứng minh rằng A=4B
\(B=1+2+2^2+...+2^6.\)
\(=>4B=2^2+2^3+...+2^8\)\(\left(1\right)\)
\(A=2^2+2^3+...+2^8\)\(\left(2\right)\)
Từ (1) và (2)
=> A = 4B
11/2mu2+1/3mu2+1/4mu2+...+1/9mu2 cmr 2/5<s<8/9
Biết rằng 1mu 2 +2mu2 +3mu2+......+10mu2= 385 tính nhanh di tổng. S=2mu2+4mu2+6mu2+.....+20mu2
S = 22 + 42 + 62 + ... + 202
= (2.1)2 + (2.2)2 + (2.3)2 ... (2.10)2
= 22.12 + 22.22 + 22.32 + ... + 22.102
= 22 (12 + 22 + ... + 102 )
= 4 . 385
= 1540
= (1x2)^2 (2x2)^2 (3x2)^2 (4x2)^2 ..... (9x2)^2 (10x2)^2
= 1^2 x 2^2 2^2 x 2^2 3^2 x 2^2 4^2 x 2^2 ..... 9^2 x 2^2 10^2 x 2^2
= (1^2 2^2 3^2 4^2 ..... 9^2 10^2) x 2^2
= 385 x 2^2 = 385 x 4 = 1540
a) (13-145+49)-(13+49) b)25*2mu2(15-18)+(12-19+10)
tính nhanh và cách làm nha bạn
a ) ( 13 - 145 + 49 ) - ( 13 + 49 )
= 13 - 145 + 49 - 13 - 49
= ( 13 - 13 ) - 145 + ( 49 - 49 )
= 0 - 145 + 0
= - 145
b) 25 . 22 . ( 15 - 18 ) + ( 12 - 19 + 10 )
= 25 . 4 . ( -3 ) + 7
= 100 . [ ( -3 ) + 7 ]
= 100 . 4
= 400
a ) ( 13 - 145 + 49 ) - ( 13 + 49 )
= 13 - 145 + 49 - 13 - 49
= ( 13 - 13 ) - 145 + ( 49 - 49 )
= 0 - 145 + 0
= - 145
( 13 - 145 + 49 ) - ( 13 + 49 )
= 13 - 145 + 49 - 13 - 49
= 13 - 13 + 49 - 49 - 145
= 0 + 0 - 145
= - 145