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a/\(A=\left(3+5\right)^2=8^2=64\)
\(B=3^2+5^2=9+25=34\)
\(\Rightarrow A>B\)
b/ \(C=\left(3+5\right)^3=8^3=512\)
\(D=3^3+5^3=27+125=152\)
\(\Rightarrow C>D\)
a/ A= (3+5)2 = 82 = 64
B = 32 + 52 = 9 + 25 = 34
vì 64>34 => A > B
b/ C = (3+5)3 = 83 = 512
D = 33 + 53 = 27 + 125 = 152
Vì 512>152 => C > D
S = 1 + 3 + 32 + 33 + ... + 38 + 39
S = ( 1 + 3 ) + ( 32 + 33 ) + ... + ( 38 + 39 )
S = 4 + ( 1 . 32 + 3 .32 ) + .. + ( 1. 38 + 3 . 38 )
S = 4 + 4 .32 + .. + 4 . 38
S = 4 ( 1 + 32 + ... + 38 ) \(⋮\)4
Vậy S \(⋮\)4 ( đpcm )
Học tốt
#Dương
S = 1 + 3 + 32 + 33 + 34+35+ 36 + 37 + 38+39
S=( 1 + 3)+(32 + 33)+(34+35)+(36 + 37)+(38+39)
s=4+32.(3+1)+32.(3+1)+34.(3+1)+36.(3+1)+38.(3+1)
S=4.(1+32+34+36+38)
CHIA HẾT CHO 4
A= 1+2+22+23+.......+298+299
A= (1+2)+(22+23)+.......+(298+299 )
A=3+22.(1+2)+...+298.(1+2)
A= 3+22.3+...+298.3
A=3.(22+...+298)
Vid 3 chia hết cho 3 nên A chia hết cho 3
Đơn giản như đang giỡn
HT
a. 52 + (x+3) = 52
=> x + 3 = 52 - 52
=> x + 3 = 0
=> x = -3
b. 23 + (x-32) = 53 - 43
=> 8 + (x-9) = 125 - 64
=> x - 9 = 125 - 64 - 8
=> x - 9 = 53
=> x = 53 + 9
=> x = 62
i) \(2345-1000\div\left[19-2\left(21-18\right)^2\right]\)
\(=\)\(2345-1000\div\left[19-2.3^2\right]\)
\(=\)\(2345-1000\div\left[19-2.9\right]\)
\(=\)\(2345-1000\div\left[19-18\right]\)
\(=\)\(2345-1000\div1\)
\(=\)\(2345-1000\)
\(=\)\(1345\)
j) \(128-\left[68+8\left(37-35\right)^2\right]\div4\)
\(=\)\(128-\left[68+8.2^2\right]\div4\)
\(=\)\(128-\left[68+8.4\right]\div4\)
\(=\)\(128-\left[68+32\right]\div4\)
\(=\)\(128-100\div4\)
\(=\)\(128-25\)
\(=\)\(3\)
k) \(568-\left\{5\left[143-\left(4-1\right)^2\right]+10\right\}\div10\)
\(=\)\(568-\left\{5\left[143-3^2\right]+10\right\}\div10\)
\(=\)\(568-\left\{5\left[143-9\right]+10\right\}\div10\)
\(=\)\(568-\left\{5.134+10\right\}\div10\)
\(=\)\(568-\left\{670+10\right\}\div10\)
\(=\)\(568-680\div10\)
\(=\)\(568-68\)
\(=\)\(500\)
a) \(107-\left\{38+\left[7.3^2-24\div6+\left(9-7\right)^3\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[7.3^2-24\div6+2^3\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[7.9-4+8\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[63-4+8\right]\right\}\div15\)
\(=\)\(107-\left\{38+67\right\}\div15\)
\(=\)\(107-105\div15\)
\(=\)\(107-7\)
\(=\)\(7\)
b) \(307-\left[\left(180-160\right)\div2^2+9\right]\div2\)
\(=\)\(307-\left[20\div4+9\right]\div2\)
\(=\)\(307-\left[5+9\right]\div2\)
\(=\)\(307-14\div2\)
\(=\)\(307-7\)
\(=\)\(300\)
c) \(205-\left[1200-\left(4^2-2.3\right)^3\right]\div40\)
\(=\)\(205-\left[1200-\left(16-6\right)^3\right]\div40\)
\(=\)\(205-\left[1200-10^3\right]\div40\)
\(=\)\(205-\left[1200-1000\right]\div40\)
\(=\)\(205-200\div40\)
\(=\)\(205-5\)
\(=\)\(200\)
a. S = 1 + 2 + 2^2 + 2^3 + ... + 2^8 + 2^9
Ta có: 2 = 1 . 2
2^2 = 2 . 2
2^3 = 2^2 . 2
.....
=> 1 + 2 + 2^2 + ... + 2^8 + (2^8 . 2)
=> 1 + 2 + 2^2 + ... + (2^8 . 3)
=> 1 + 2 + 2^2 + ... + 2^7 + (2^7 .6)
=> 1 + 2 + 2^2 + ... + (2^7 . 7)
=> .....
=> 1 + 2 . 311
a/ A = 82
A = 64
B = 9 + 25, B = 34
b/ C = 83, CC = 512
D = 27 + 125
D = 152