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Câu 1,
\(S=1+2+2^2+...+2^7\)
\(=\left(1+2\right)+2^2\left(1+2\right)+2^4\left(1+2\right)+2^6\left(1+2\right)\)
\(=3+2^2.3+2^4.3+2^6.3\)
\(=3\left(1+2^2+2^4+2^6\right)⋮3\)
Nên S chia hết cho 3
Câu 2 ,
\(A=5+5^2+5^3+...+5^{20}\)
\(=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{19}\left(1+5\right)\)
\(=5.6+5^3.6+...+5^{19}.6\)
\(=6\left(5+5^3+...+5^{19}\right)⋮6\)
Nên A chia hết cho 6
số sh cua tong A bang so hang cua day so cach deu 1 don vi tu 1 den 60
so sh cua tong A la:(60-1):1+1=60 (sh)
Cu 3 sh lien tiep cua tong A nhom thanh 1 nhom thi ta duoc so nhom la : 60: 3=20(nhom)
khi do : A = (2+2^2+2^3)+(2^4+2^5+2^6)+(2^7+2^8+2^9)+....+(2^58+2^59+2^60)
A=(2+2.2+2.2^2)+(2^4+2^4.2+2^4.2^2)+(2^7+2^7.2+2^7.2^2)+.....+(2^58
2^58.2+2^58.2^2)
A=2(1+2+2^2)+2^4(1+2+2^2)+2^7(1+2+2^2)+...+2^58(1+2+2^2)
A=2.7+2^4.7+2^7.7+...+2^58.7
A=7(2+2^4+2^7+...+2^58)
Vi 7 chia het cho 7
2+2^4+2^7+...+2^58 thuoc N
Suy ra 7(2+2^4+2^7+...+2^58) chia het cho 7
hay A chia het cho 7
Vay A chia het cho 7
Câu 1:
abc >/ 100 ; bca >/ 100 ; cab>/100
< = > abc + bca + cab >/300
< = > abc + bca + cab >/ 111
b) 230 và 320
Ta có :
230 = ( 23 )10 = 810
320 = ( 32 )10 = 910
Vì 8 < 9 Nên 230 < 320
c) 1020 và 9010
Ta có :
1020 = ( 102 )10 = 10010
Vì 10010 > 9010
Nên 1020 > 9010
trả lời
bài này dễ
bn tự làm nha
chúc bn thành công trong học tập
+) \(A=3\left(x-4\right)^4-4\ge-4\)
Min A = -4 \(\Leftrightarrow x-4=0\Leftrightarrow x=4\)
+) \(B=5+2\left(x-2019\right)^{2020}\ge5\)
Min B = 5 \(\Leftrightarrow x-2019=0\Leftrightarrow x=2019\)
+) \(C=5+2018\left(2020-x\right)^2\)
Min C = 5 \(\Leftrightarrow2020-x=0\Leftrightarrow x=2020\)
+) \(D=\left(x-1\right)^{2020}+\left(y+x\right)-1\ge-1\)
Min D = -1 \(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-x\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=-1\end{cases}}}\)
+) \(E=2\left(x-1\right)^2+3\left(2x-y\right)^4-2\ge-2\)
Min E = -2 \(\Leftrightarrow\hept{\begin{cases}x-1=0\\2x-y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\2x=y\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\end{cases}}}\)
a, A = 1 + 2 + 22 + ... + 299
= (1 + 2) + (22 + 23) + ... + (298 + 299)
= 1(1 + 2) + 22(1 + 2) + ... + 298(1 + 2)
= 1 . 3 + 22 . 3 + ... + 298 . 3
Vì 3 chia hết cho 3 nên 1 . 3 + 22 . 3 + ... + 298 . 3 chia hết cho 3
hay A chia hết cho 3 (đpcm)
b, A = 1 + 2 + 22 + ... + 299
= (1 + 2 + 22 + 23) + (24 + 25 + 26 + 27) + ... + (296 + 297 + 298 + 299)
= 1 . 15 + 24 . 15 + ... + 296 . 15
Vì 15 chia hết cho 15 nên 1 . 15 + 24 . 15 + ... + 296 . 15 chia hết cho 15
hay A chia hết cho 15 (đpcm)
Tiếp bài của @trankhanhvy2008
A = 1 + 2 + 22 + 23 + 24 + ... + 299
2A = 2( 1 + 2 + 22 + 23 + 24 + ... + 299 )
= 2 + 22 + 23 + 24 + ... + 2100
2A - A = ( 2 + 22 + 23 + 24 + ... + 2100 ) - ( 1 + 2 + 22 + 23 + 24 + ... + 299 )
=> A = 2 + 22 + 23 + 24 + ... + 2100 - 1 - 2 - 22 - 23 - 24 - ... - 299
= 2100 - 1
2100 - 1 < 2100
=> A < 2100
Đề: X=\(\frac{1}{1+2}\)+\(\frac{1}{1+2+3}\)+.......+\(\frac{1}{1+2+3+4+20}\)
X=\(\frac{1}{2.3:2}\)+\(\frac{1}{3.4:2}\)+\(\frac{1}{4.5:2}\)+......+\(\frac{1}{20.21:2}\)
X=\(\frac{2}{2.3}\)+\(\frac{2}{3.4}\)\(\frac{2}{4.5}\)+........+\(\frac{2}{20.21}\)
X=2.(\(\frac{1}{2}\).3+\(\frac{1}{3}\).4+\(\frac{1}{4}\).5+.....+\(\frac{1}{20}\).21)
X=2.(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+......+\(\frac{1}{20}\)-\(\frac{1}{21}\))
X=2.(\(\frac{1}{2}\)-\(\frac{1}{21}\))
X=2.(\(\frac{21}{42}\)-\(\frac{2}{42}\))
X=2.\(\frac{19}{42}\)
X=\(\frac{19}{21}\)
Mn xem thử đúng ko nha!
Ta có: \(1+2=\frac{2.3}{2}\); \(1+2+3=\frac{3.4}{2}\); .......... ; \(1+2+3+....+20=\frac{20.21}{2}\)
\(\Rightarrow X=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+.......+\frac{1}{\frac{20.21}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+........+\frac{2}{20.21}=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{20.21}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{20}-\frac{1}{21}\right)=2.\left(\frac{1}{2}-\frac{1}{21}\right)=2.\frac{19}{42}=\frac{19}{21}\)
Merry Christmas
\(A=2^0+2^1+2^2+...+2^{19}-2^{20}\)
Dat \(M=2^0+2^1+2^2+...+2^{19}\)
\(\Rightarrow2M=2^1+2^2+2^3+...+2^{20}\)
\(\Rightarrow M=2M-M=2^{20}-1\)
\(\Rightarrow A=M-2^{20}=\left(2^{20}-1\right)-2^{20}=-1\)