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a)3^200=3^2.100=9^100
2^300=2^3.100=8^100
Suy ra 3^200>2^300
b)125^5=(5^3)^5=5^15
25^7=(5^2)7=5^14
Suy ra 125^5>25^7
c)9^20=(3^2)^20=3^40
27^13=(3^3)13=3^39
Suy ra 9^20>27^13
d)3^54=3^6.9=(3^6)^27=729^9
2^81=2^9.9=512^9
Suy ra 3^54>2^81
e)10^30=10^3.10=1000^10
2^100=2^10.10=1024^10
Suy ra 10^30<2^100
f)5^40=5^4.10=625^10
Suy ra 5^40>620^10
a) \(63^7< 64^7=\left(2^6\right)^7=2^{42}< 2^{48}=\left(2^4\right)^{12}=16^{12}\Rightarrow63^7< 16^{12}\)
b) \(3^{151}>3^{150}=\left(3^2\right)^{75}=9^{75}>8^{75}=\left(2^3\right)^{75}=2^{225}\)
c) \(9^{20}=\left(3^2\right)^{20}=3^{40}>3^{39}=\left(3^3\right)^{13}=27^{13}\Rightarrow9^{20}>27^{13}\)
bài 2:
a)\(2^x=32\Leftrightarrow2^x=2^5\Leftrightarrow x=5\)
b)\(2x+3^4=7^2\Leftrightarrow2x+81=49\Leftrightarrow2x=-32\Leftrightarrow x=-16\)
c)\(12x-33=3^2\Leftrightarrow12x-33=9\Leftrightarrow12x=42\Leftrightarrow x=\frac{7}{2}\)
A=5+52+53+....+59+510
=> A=(5+52)+(53+54)+...+(59+510)
=> A=5(1+5)+53(1+5)+....+59(1+5)
=> A=5.6+53.6+....+59.6
=> A=6(5+53+....+59)
=> A chia hết cho 6 (đpcm)
a) Tổng S có 100 số hạng chia thành 25 nhóm , mỗi nhóm có 4 số hạng :
\(S=1-3+3^2-3^3+...+3^{98}-3^{99}\)
\(S=\left(1-3+3^2-3^3\right)+\left(3^4-3^5+3^6-3^7\right)+...+\left(3^{96}-3^{97}+3^{98}-3^{99}\right)\)
\(S=-20+3^4.\left(-20\right)+...+3^{96}\left(-20\right)⋮-20\)
b)\(S=1-3+3^2-3^3+...+3^{98}-3^{99}\)
\(\Leftrightarrow3S=3-3^2+3^3-3^4+...+3^{99}-3^{100}\)
Cộng từng vế của 2 đẳng thức ta có :
\(3S+S=\left(3+1\right)S=4S=\frac{1-3^{100}}{4}\)
Vì S là 1 số nguyên nên 1 - 3100 chia hết cho 4 hay 3100 -1 chia hết cho 4 => 3100 chia 4 dư 1
a)\(2^x.4=128\Leftrightarrow2^x=32\Leftrightarrow2^x=2^5\Rightarrow x=5\)
b)\(\left(2x+1\right)=125\Leftrightarrow2x=126\Leftrightarrow x=13\)
c)\(x^{15}=x\Leftrightarrow\orbr{\begin{cases}x=\pm1\\x=0\end{cases}}\)
d) \(\left(x-5\right)^4=\left(x-5\right)^5\Leftrightarrow\orbr{\begin{cases}x-5=1\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=5\end{cases}}\)
a,
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
b,
2x = 124
x = 62
c,
\(x^{15}-x=0\)
\(x\left(x^{14}-1\right)=0\)
\(\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x^{14}=1\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
d,
\(0=\left(x-5\right)^5-\left(x-5\right)^4\)
\(\left(x-5\right)^4\left(x-5-1\right)=0\)
\(\orbr{\begin{cases}\left(x-5\right)^4=0\\x-6=0\end{cases}}\)
\(\orbr{\begin{cases}x=5\\x=6\end{cases}}\)
5^6+5^7+5^8
=5^6.(1+5+5^2)
=5^6.31 chia hết cho 31
7^6+7^5-7^4
=7^4.(7^2+7-1)
=7^4.55 chia hết cho 11
BÀI 2:
a) \(5^6+5^7+5^8=5^6\left(1+5+5^2\right)=5^6.31\) \(⋮\)\(31\)
b) \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.55\)\(⋮\)\(11\)
c) \(2^3+2^4+2^5=2^3.\left(1+2+2^2\right)=2^3.7\)\(⋮\)\(7\)
d) mk chỉnh đề
\(1+2+2^2+2^3+...+2^{59}\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{58}+2^{59}\right)\)
\(=\left(1+2\right)+2^2\left(1+2\right)+...+2^{58}\left(1+2\right)\)
\(=\left(1+2\right)\left(1+2^2+...+2^{58}\right)\)
\(=3\left(1+2^2+...+2^{58}\right)\)\(⋮\)\(3\)
Trả lời :
a, 59 - x = 125
=> 59 - x = 52
=> 9 - x = 2
=> x = 7
b,3x : 27 = 9
=> 3x : 33 = 32
=> x = 2 + 3
=> x = 5
c, 2.x3 - 4 = 12
=> 2 . x3 = 16
=> x3 = 8
=> x3 = 23
=> x = 2
d, 2 . x -33 = 125
=> 2x - 27 = 125
=> 2x = 152
=> x = 76
a, \(5^{9-x}=125\)
\(\Rightarrow5^{9-x}=5^3\)
\(\Rightarrow9-x=3\)
\(\Rightarrow x=9-3=6\)
Vậy x = 6
b, \(3^{\text{x}}:27=9\)
\(\Rightarrow3^{\text{x}}\div3^3=3^2\)
\(\Rightarrow x=5\)
Vậy x = 5
c, \(2.x^3-4=12\)
\(\Rightarrow2\text{x}^3=16\)
\(\Rightarrow x^3=8\)
\(\Rightarrow x=2\)
Vậy x = 2
d, \(2\text{x}-3^3=125\)
\(\Rightarrow2\text{x}-27=125\)
\(\Rightarrow2\text{x}=125+27\)
\(\Rightarrow2\text{x}=152\)
\(\Rightarrow x=\frac{152}{2}=76\)
Vậy x =76
(8110- 273-921):225
=(34)10-(33)3-(32)21
=340-39-342
=339.(-25)
=-225.337
-225 nhân cho 1 số tự nhiên thì luôn luôn chia hết cho 225 nhoa