20+ 19+ 18+ ...+ x = 20
Tích này tận cùng có bao nhiêu chữ số 0?
-20 x -19 x -18......18 x 19 x 20
Chúng đã nhân với 0 nên chữ số tận cùng có 1 chữ số 0, đó là 0
Tính nhanh:
11 x 12 + 12 x 13 + 13 x 14 + 14 x 15 + 15 x 16 + 16 x 17 + 17 x 18 + 18 x 19 + 19 x 20 + 20 x 21
11 x 12 + 12 x 13 + 13 x 14 + 14 x 15 + 15 x 16 + 16 x 17 + 17 x 18 + 18 x 19 + 19 x 20 + 20 x 21
= 12 x ( 11 + 13 ) + 14 x ( 13 + 15 ) + 16 x ( 15 + 17 ) + 18 x ( 17 + 19 ) + 20 x 21
= 12 x 24 + 14 x 28 + 16 x 32 + 18 x 36 + 420
= 288 + 392 + 512 + 648 + 420
= ( 288 + 392 ) + ( 512 + 648 ) + 420
= 680 + 1160 + 420
= 1840 + 420
= 2260
tk nha
=
Ta có:
11 x 12 + 12 x 13 + 13 x 14 + 14 x 15 + 15 x 16 + 16 x 17 + 17 x 18 + 18 x 19 + 19 x 20 + 20 x 21
= 12 x ( 11 + 13 ) + 14 x ( 13 + 15 ) + 16 x ( 15 + 17 ) + 18 x ( 17 + 19 ) + 20 x 21
= 12 x 24 + 14 x 28 + 16 x 32 + 18 x 36 + 420
= 288 + 392 + 512 + 648 + 420
= ( 288 + 392 ) + ( 512 + 648 ) + 420
= 680 + 1160 + 420
= 1840 + 420
= 2260
#Mạt Mạt#
11 x 12 + 12 x 13 + 13 x 14 + 14 x 15 + 15 x 16 + 16 x 17 + 17 x 18 + 18 x 19 + 19 x 20 + 20 x 21
= 12 x ( 11 + 13 ) + 14 x ( 13 + 15 ) + 16 x ( 15 + 17 ) + 18 x ( 17 + 19 ) + 20 x ( 19 + 21 )
= 12 x 24 + 14 x 28 + 16 x 32 + 18 x 36 + 20 x 40
= 288 + 392 + 512 + 648 + 800
= ( 288 + 392 ) + ( 512 + 648 ) + 800
= 680 + 1160 + 800
= 2640
Tìm x thuộc Z biết:
20 = 20 + 19 + 18 +...+ 1 + x
Đặt A = 1 + 2 + 3 + ... + 20
Ssh của A là : ( 20 - 1 ) : 1 + 1 = 20
A = ( 20 + 1 ) . 20 : 2 = 210
Ta có : 20 = 210 + x <=> = 20 - 210 => x = - 190
Vậy x = - 190
( 10+12+13+...+18+19+20)x(20-12-8)
( 10 + 12 + 13 + .... + 18 + 19 + 20 ) x ( 20 - 12 - 8 )
= ( 10 + 12 + 13 + .... + 18 + 19 + 20 ) x ( 8 - 8 )
= ( 10 + 12 + 13 + ... + 18 + 19 + 20 ) x 0
= 0
=(10+20)+(12+18)+(13+17)+(14+16)+(15+19) .<20-(12+8)>
= 30+30+30+30+30.(20-20)
= 150.0
=0
\(ChoM=\frac{18}{18+19+20}+\frac{19}{18+19+21}+\frac{20}{18+19+21}+\frac{21}{18+19+21}.\)
Chứng tỏ rằng 1<M<2
Có \(\frac{18}{18+19+20}>\frac{18}{18+19+20+21}\)
\(\frac{19}{18+19+21}>\frac{19}{18+19+20+21}\)
\(\frac{20}{18+19+21}>\frac{20}{18+19+20+21}\)
\(\frac{21}{18+19+21}>\frac{21}{18+19+20+21}\)
=> \(\frac{18}{18+19+20}+\frac{19}{18+19+21}+\frac{20}{18+19+21}+\frac{21}{18+19+21}>\frac{18}{18+19+20+21}+\frac{19}{18+19+20+21}+\frac{20}{18+19+20+21}+\frac{21}{18+19+20+21}\)
=> \(\frac{18}{18+19+20}+\frac{19}{18+19+21}+\frac{20}{18+19+21}+\frac{21}{18+19+21}>\frac{18+19+20+21}{18+19+20+21}\)
=>\(\frac{18}{18+19+20}+\frac{19}{18+19+21}+\frac{20}{18+19+21}+\frac{21}{18+19+21}>1\)
=>M>1
Còn lại mình không biết, đúng thì tick nha
18+19+20+21+123456+x+x=
18+19+20+21+123456+x+1
=123534+x+1
coi x la 1 ta co;
123534+x+1
=123534+1+1
=123535+1
=123536
A=20^18+1/20^19+1,B=20^19+1/20^20+1.Hãy so sánh A và B
\(B=\dfrac{20^{19}+1}{20^{20}+1}< \dfrac{20^{19}+1+19}{20^{20}+1+19}=\dfrac{20^{19}+20}{20^{20}+20}\)
\(B< \dfrac{20.\left(20^{18}+1\right)}{20.\left(20^{19}+1\right)}\)
\(B< \dfrac{20^{18}+1}{20^{19}+1}\)
\(B< A\)
(x-20)+(x-19)+(x-18)+.....+100+101=101
(x-20)+(x-19)+(x-18)+.....+100+101=101
(x-20)+(x-19)+(x-18)+.....+100 =0
\(\frac{\left(x-20+100\right)n}{2}\) =0
(x+80)n =0
x+80 =0
x =0-80
x =-80
Vậy x=-80
(x-20)+(x-19)+(x-18)+...+100+101=101