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Bài 2:
Ta có \(1^2+3^2+6^2+...+36^2\)
\(=1^2+3^2+\left(3.2\right)^2+...+\left(3.12\right)^2\)
\(=1^2+3^2+3^2.2^2+...+3^2.12^2\)
\(=1^2+3^2+3^2\left(2^2+3^2+...+12^2\right)\left(1\right)\)
Mà \(2^2+3^2+...+12^2=649\)
Nên \(\left(1\right)=1+9+9.649\)
\(=10+5841=5851\)
\(\Rightarrow1^2+3^2+6^2+...+36^2=5851\)
bai 2: a) \(2^{30}=\left(2^3\right)^{10}=8^{10}\)
\(3^{20}=\left(3^2\right)^{10}=9^{10}\)
vi 810 <910 nen 230 <320
b) \(5^{202}=\left(5^2\right)^{101}=25^{101}\)
\(2^{505}=\left(2^5\right)^{101}=32^{101}\)
vi 25101 <32101 nen 5202 <2505
c) \(333^{444}=\left(3.111\right)^{444}=3^{444}.111^{444}=\left(3^4\right)^{111}.111^{444}=81^{111}.111^{444}\)
\(444^{333}=\left(4.111\right)^{333}=4^{333}.111^{333}=\left(4^3\right)^{111}.111^{333}=64^{111}.111^{333}\)
vi 81111>64111 va 111444>111333
nen 333444>444333
bai 3 : \(\left(\frac{1}{3}\right)^{2n-1}=3^5\)
\(\left(\frac{1}{3}\right)^{2n-1}=\left(\frac{1}{3}\right)^{-5}\)
2n-1=-5
2n=-5+1
2n=-4
n=-4:2
n=-2
Bai 4 : 3x-5/9=0 va 3y+0,4/3=0
3x=5/9 va 3y=2/15
x=5/27 va y=2/45
Bai 5:
A=75. {42002.(42+1)+....+(42+1)+1)+25
A=75.{42002.20+...+20+1}+25
A=75.{20.(42002+...+1)+1}+25
A=75.20.(42002+..+1)+75+25
A=1500.(42002+...+1)+100
A=100.{15.(42002+...+1)+1} chia het cho 100
\(10^{30}=\left(10^3\right)^{10}=1000^{10}\)
\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)
Vì \(1000^{10}< 1024^{10}\Rightarrow10^{30}< 2^{100}\)
a, Ta có:
1030 = (103)10 = 100010
2100 = (210)10 = 102410
Vì 1000 < 1024
=> 100010 < 102410
hay 1030 < 2100
b, Ta có:
222555 = (2225)111 = (1115.25)111
= (1115 . 32)111
Lại có:
555222 = (5552)111 = (1112 . 52)111
= (1112 . 25)111
Ta có:
1112 < 1115
=> 1112.25 < 1115 . 32
=>(1112 . 25)111 < (1112 . 25)111
hay 555222 < 222555
222555 = ( 2.111 )5.111 = 25.111.1115.111
555222 = ( 5.111 )2.111 = 52.111 .1112.111
Vì 25 > 52 ( 32 > 25 ) và 1115 > 1112 ( 5 > 2 ) nên 25.111.1115.111 > 52.111 .1112.111
hay 222555 > 555222
\(\frac{222^{555}}{555^{222}}=\frac{\left(2.111\right)^{\left(5.111\right)}}{\left(5.111\right)^{\left(2.111\right)}}=111^{\left(111\left(5-2\right)\right)}.\left(\frac{2^5}{5^2}\right)^{111}=111^{333}.\left(\frac{32}{25}\right)^{1111}>1\)
\(222^{555}>555^{222}\)
Bài 1:
a) \(\frac{\left(-3\right)^n}{81}=9\Leftrightarrow\left(-3\right)^n=9.81=729\Rightarrow\left(-3\right)^n=\left(-3\right)^6\Rightarrow n=6\)
b) \(\frac{125}{5^n}=5^2\Leftrightarrow\frac{125}{5^n}=25\Rightarrow5^n=125:25=5\Rightarrow n=1\)
Bài 2:
a) \(625^5=\left(5^4\right)^5=5^{4.5}=5^{20}\)
\(125^7=\left(5^3\right)^7=5^{3.7}=5^{21}\)
Thấy: \(5^{20}< 5^{21}\Rightarrow625^5< 125^7\)
b) \(3^{2n}=\left(3^2\right)^n=9^n\) ; \(2^{3n}=\left(2^3\right)^n=8^n\)
\(9^n>8^n\Rightarrow3^{2n}>2^{3n}\)
K cho mình nhé.
\(\left\{{}\begin{matrix}ac=b^2\Rightarrow\frac{a}{b}=\frac{b}{c}\\ab=c^2\Rightarrow\frac{b}{c}=\frac{c}{a}\end{matrix}\right.\) \(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\Rightarrow a=b=c\)
\(\Rightarrow P=1+1+1=3\)
ta có: 222555 = ( 2.111)555 = 2555.111555= (25)111.111555 = 32111.111555
555222 = ( 5.111)222 = 5222.111222 = ( 52)111.111222 = 25111.111222
mà 32111> 25111; 111555>111222
=> 32111.111555>25111.111222
=> 222555> 555222
Ta có:
222555 = 111555.2555 = 111222.111333.(25)111 = 111222.111333.32111
555222 = 111222.5222 = 111222.(52)111 = 111222.25111
Do 111333.32111 > 25111
=> 111222.111333.32111 > 111222.25111
=> 222555 > 555222
222555=(2225)111
555222=(5552)111
Đến đây, ta chỉ cần so sánh 2225 và 5552
Ta có 2222>555 ( vì 2222 nhiều chữ số hơn
Do đó (2222)2>5552
=>2224>5552
=>2225>5552
Vậy 222555>555222