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Bài 1 :
a) Ta có : 3210 = (25)10 = 250
1615 = (24)15 = 260
250 < 260 => 3210 < 1615
b) Ta có : 2711 = (33)11 = 333
818 = (34)8 = 332
333 > 332 => 2711 > 818
c) Ta có : 536 = (53)12 = 12512
1124 = (112)12 = 12112
12512 > 12112 => 536 > 1124
d) Ta có : 216 = 213 . 2 . 2 . 2 = 213 . 8
7. 213 < 213 . 8 => 7 . 213 < 216
Bài 3 :
Ta có :
S = 1 + 2 + 22 + 23 + ... + 22018
S = (1 + 2) + (22 + 23 + 24) + ... + (22016 + 22017 + 22018)
S = 3 + 28 + ... + 22015(2 + 22 + 23)
S = 3 + 28 + ... + 22015. 14
Vậy số dư khi chia S cho 7 là 3
a, \(A=2015.20162016-2016.20152015\)
\(A=2015.\left(2016.10001\right)-2016.20152015\)
\(A=\left(2015.10001\right).2016-20152015.2016\)
\(A=20152015.2016-20152015.2016\)
\(A=0\)
Vậy A = 0
b, \(B=\left(3.4.2^{16}\right)^2\div11.2^{13}.4^{11}-16^9\)
\(B=3^2.2^4.2^{32}\div11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9\)
\(B=3^2.2^4.2^{32}\div11.2^{13}.2^{22}-2^{36}\)
\(B=3^2.2^{36}\div11.2^{35}-2^{36}\)
\(B=3^2.2^{35}.2\div11.2^{35}-2.2^{35}\)
\(B=3^2.2\div9=9.2\div9=2\)
Vậy B = 2
c, \(C=2^{10}.13+2^{10}.65\div2^8.104\)
\(C=2^{10}.\left(13+65\right)\div2^8.104\)
\(C=2^{10}.78\div2^8.104\)
\(C=2^{10}.39\div2^8.13\)
\(C=39\div13=3\)
Vậy C = 3
Đề bài câu c sai mk sửa nhé là 28 ms tính đc k nó dư lắm !!!
Bài 1:
A,
(100+121+144)÷(169+196)
=365:365=1
B
,1.2.3...7.8.(9-1-8)
=1.2.3...7.8.0=0
C,
3^2.2^4.2^32/11.2^13.2^22-2^26
=3^2.2^36/11.2^35-2^36
=3^2.2^36/2^35-(11-2)
=9.2/9=2
D,
=1152-374-1152+(-65)+374
=(1152-1152)+(-374+374)+(-65)
=0+0+(-65)=-65
E,
=13-(12-11-10+9)+(8-7-6+5)-(4-3-2+1)
=13-0+0-0=13
Bài 2:
A,(19x+50)÷14=25-16
(19x+50)÷14=9
19x+50=126
19x=76
x=4
B,
31x+(1+2+3+...+30)=1240
31x+465=1240
31x=775
×=25
a)\(\left(10^2+11^2+12^2\right)\div\left(13^2+14^2\right)\)
\(=\left(100+121+144\right)\div\left(169+196\right)\)
\(=365\div365\)
\(=1\)
b) \(1.2.3...9-1.2.3...8-1.2.3...8^2\)
\(=1.2.3...8\left(9-1-8\right)\)
\(=1.2.3...8.0\)
\(=0\)
d) \(1152-\left(374+1152\right)+\left(-65+374\right)\)
\(=1152-374-1152-65+374\)
\(=\left(1152-1152\right)-65+\left(374-374\right)\)
\(=0-65+0\)
\(=-65\)
e) \(13-12+11+10-9+8-7-6+5-4+3+2-1\)
\(=13-\left(12-11\right)+\left(10-9\right)+\left(8-7\right)-\left(6-5\right)-\left(4-3\right)\)\(+\left(2-1\right)\)
\(=13-1+1+1-1-1+1\)
\(=13+0+0+0\)
\(=13\)
CÂU 1
\(5^{n+1}+5^n=750\)
\(=>5^n\cdot5+5^n=750\)
\(=>5^n\cdot\left(5+1\right)=750\)
\(=>5^n\cdot6=750\)
\(=>5^n=750:6\)
\(=>5^n=125\)
\(=>5^n=5^3\)
\(=>n=3\)
b10:
1.\(A=\left(\frac{999-1}{2}+1\right).\frac{999+1}{2}=250000\)
2. \(B=\left(1+3+...+2017\right)-\left(2+4+...+2016\right)\)
\(=2017.\frac{2017+1}{2}-\left(\frac{2016-2}{2}+1\right).\frac{2016+2}{2}\)
đến đây bạn bấm máy đi nhé!
3. \(C=3+3^2+3^3+...+3^{99}\left(1\right)\)
Nhân hai vế của (1) vs số 3 ta được:
\(3C=3^2+3^3+...+3^{100}\left(2\right)\)
Lấy (2)-(1) theo vế ta được: \(3C-C=3^{100}-3\)
=> C=\(\frac{3^{100}-3}{2}\)
4. Làm giống hết câu 3 luôn nhé, chỉ là nhân với 4 thôi.
1/ tính :
a/ A = 341 . 67 + 341 . 16 + 659 . 83
A = 341 . ( 67 + 16 ) + 659 . 83
A = 341 . 83 + 659 . 83
A = 83 . ( 341 + 659 )
A = 83 . 1000
A = 83 000
b/ B = 42 . 53 + 47 . 156 - 47 . 114
B = 42 . 53 + 47 . ( 156 - 114 )
B = 42 . 53 + 47 . 42
B = 42 . ( 53 + 47 )
B = 42 . 100
B = 4 200
2/ thu gọn tổng :
A = 3 + 32 + 33 + ... + 3100
3A = 3^2 + 3^3 + 3^4 + ...+ 3^101
3A - A = ( 3^2 + 3^3 + 3^4 + ...+ 3^101 ) - ( 3 + 32 + 33 + ... + 3100 )
2A = 3^101 - 3
A = 3^101 - 3 / 2
Bài 1:
\(A=341.67+341.16+659.83.\)
\(=341.\left(67+16\right)+659.83\)
\(=341.83+659.83\)
\(=83.\left(341+659\right)\)
\(=83.1000=83000\)
\(B=42.53+47.156-47.114\)
\(=42.53+47.\left(156-114\right)\)
\(=42.53+47.42\)
\(=42.\left(47+53\right)\)
\(=42.100=4200\)
Bài 2:
\(A=3+3^2+3^3+3^4+....+3^{100}\)
\(\Rightarrow3A=3^2+3^3+3^4+3^5+...+3^{101}\)
\(2A=3A-A=\left(3^2+3^3+3^4+3^5+....+3^{101}\right)-\left(3+3^2+3^3+3^4+....+3^{100}\right)\)
\(\Rightarrow2A=3^{101}-3\)
\(\Rightarrow A=\frac{3^{101}-3}{2}\)
Bài 3:
\(S=1+2+3+4+...+2018\)
\(=\frac{\left[1+2018\right].\left[\left(2018-1\right)+1\right]}{2}\)
\(=\frac{2019.2018}{2}=2037171\)
\(P=1+3+5+7+...+2017\)
\(=\frac{\left[2017+1\right].\left[\left(2017-1\right):2+1\right]}{2}\)
\(=\frac{2018.1009}{2}\)
\(=1018081\)
Bài 4:
\(\text{Vì ab = 0 }\)\(\Rightarrow\)\(a=0\)\(\text{hoặc}\)\(b=0\)
\(\text{Th1 : ( a = 0)}\)
\(a+4b=16\)
\(0+4b=16\)
\(4b=16\Leftrightarrow b=4\)
\(\text{Th2: ( b = 0)}\)
\(a+4b=16\)
\(a+4.0=16\)
\(a+0=16\Leftrightarrow a=16\)
\(\text{Vậy :}\)\(a;b\in\left\{0;4\right\};\left\{16;0\right\}\)
Bài 5:
\(A=\frac{10^2+11^2+12^2}{13^2+14^2}=\frac{365}{365}=1\)
\(B=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{\left(12.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{12^2.2^{32}}{11.2^{13}.4^{11}-16^9}=....=2\)