2225 = 23.75 = (23)75 = 875

3150 = 32.75 = (32)75=975

8 < 9 ⇒ 875 < 975

Vậy : 2225 < 3150

a/ \(2^{225}=\left(2^3\right)^{75}=8^{75}\)

\(3^{151}>3^{150}=\left(3^2\right)^{75}=9^{75}\)

Mà \(8^{75}< 9^{75}\)

=> \(2^{225}< 3^{150}< 3^{151}\)

b/ Xét n là số lẻ

=> n + 1 chẵn

=> n + 1 ⋮ 2

=> (n+1)(3n+2) ⋮2

Xét n là số chẵn

=> 3n chẵn

=> 3n+2 chẵn

=> (n+1)(3n+2) ⋮2

Do đó A = (n+1)(3n+2) chia hết cho 2 với mọi số tự nhiên n 

Bạn ấn\(fx\)là ra phân số mà.

Mình giải như sau:

\(\frac{228}{225}>\frac{2228}{2225}\)

\(\frac{36}{39}>\frac{35}{41}\)

nhé bạn

\(\frac{228}{225}>\frac{2228}{2225}\)

\(\frac{36}{39}>\frac{35}{41}\)

Bạn giải thế nào ?

\(1,\\ a,2^x=16=2^4\Rightarrow x=4\\ b,3^{x+1}=9^x=3^{2x}\\ \Rightarrow x+1=2x\Rightarrow x=1\\ c,2^{3x+2}=4^{x+5}=2^{2\left(x+5\right)}\\ \Rightarrow3x+2=2x+10\Rightarrow x=8\\ d,3^{2x-1}=243=3^5\\ \Rightarrow2x-1=5\Rightarrow x=3\\ 2,\\ a,2^{225}=8^{75}< 9^{75}=3^{150}\\ b,2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=\left(5^5\right)^7=5^{35}\\ c,99^{20}=\left(99^2\right)^{10}< \left(99\cdot101\right)^{10}=9999^{10}\\ 3,\\ a,12^8\cdot9^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}=\left(2\cdot3^2\right)^{16}=18^{16}\\ b,75^{20}=\left(3\cdot5^2\right)^{20}=3^{20}\cdot5^{40}=\left(3^{20}\cdot5^{10}\right)\cdot5^{30}=\left(3^2\cdot5\right)^{10}\cdot5^{30}=45^{10}\cdot5^{30}\)

Bài 1:

a) \(\Rightarrow2^x=2^4\Rightarrow x=4\)

b) \(\Rightarrow3^{x+1}=3^{2x}\Rightarrow x+1=2x\Rightarrow x=1\)

c) \(\Rightarrow2^{3x+2}=2^{2x+10}\Rightarrow3x+2=2x+10\Rightarrow x=8\)

d) \(\Rightarrow3^{2x-1}=3^5\Rightarrow2x-1=5\Rightarrow x=3\)

Bài 2:

a) \(2^{225}=\left(2^3\right)^{75}=8^{75}< 9^{75}=\left(3^2\right)^{75}=3^{150}\)

b) \(2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=\left(5^5\right)^7=5^{35}\)

c) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\)

Bài 3:

a) \(12^8.9^{12}=\left(4.3\right)^8.9^{12}=4^8.3^8.9^{12}=2^{16}.9^4.9^{12}=2^{16}.9^{16}=\left(2.9\right)^{16}=18^{16}\)

b) \(75^{20}=\left(75^2\right)^{10}=5625^{10}=\left(45.125\right)^{10}=45^{10}.125^{10}=45^{10}.5^{30}\)

Vì \(\frac{-1}{25}0\)

=>\(\frac{-1}{25}

\(\frac{201201}{203203}

Bài 8:

a) \(2^{225}=\left(2^3\right)^{75}=8^{75}\)

\(3^{150}=\left(3^2\right)^{75}=9^{75}\)

Vì \(8^{75}< 9^{75}\Rightarrow2^{225}< 3^{150}\)

b) \(2^{91}=\left(2^{13}\right)^7=8192^7\)

\(5^{35}=\left(5^5\right)^7=3125^7\)

Vì \(8192^7>3125^7\Rightarrow2^{91}>5^{35}\)

c) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\)

\(A=2219.2221.2226-2218.2223.2225\)

\(A=2219.(2223-2).2226-2218.2223.2225\)

\(A=2219.2223.2226-2.2219.2226-2218.2223.2225\)

\(A=2223.(2219.2216-2218.2225)-2.2219.2216\)

\(A=2223.\left\{(2218+1).(2215+1)-2218.2225\right\}-2.2219.2216\)

\(A=2223.(2218+2225+1)-2219.2226-2219.2226\)

\(A=2223.2219+2223.2225-2219.2226-2219.2226\)

\(A=(2223.2219-2219.2226)+2223.2225-2219.2225-2219\)

\(A=2219.(-3)+2225.4-2219\)

\(A=2219.(-4)+2225.4\)

\(A=4.(2225-2219)\)

\(A=4.6\)

\(A=24\)

\(B=3004.2999.2997-3003.2996.3001\)

\(B=3004.2999.(3001-4)-3003.2996.3001\)

\(B=(3003+1).2999.3001-3004.2999.4-3003.2996.3001\)

\(B=3003.(2996+3).3001+2999.3001-3004.2999.4-3003.2996.3001\)

\(B=3003.2996.3001+3.3003.3001-3.3004.2999+2999.3001-3004.2999-3003.2996.3001\)

\(B=3.(3003.3001-3004.2999)+2999.(3001-3004)\)

\(B=3.\left\{\left(3004-1\right).\left(2999+2\right)-3004.2999\right\}-3.2999\)

\(B=3.\left(3004.2999+2.3004-2999-2-3004.2999\right)-3.2999\)

\(B=3.(2.3003-2999)-3.2999\)

\(B=6.3003-6.2999\)

\(B=6.(3003-2999)\)

\(B=6.4\)

\(B=24\)

Mà \(A=24\) , \(B=24\)

\(\Rightarrow A=B\)

A=2219*(2223-2)*2226 - 2218*2223*2225

=2219*2223*2226 - 2*2219*2226 - 2218*2223*2225

=2223*(2219*2226 - 2218*2225) - 2*2219*2226

=2223*[(2218+1)*(2225+1) - 2218*2225] - 2*2219*2226

=2223*(2218+2215+1) - 2*2219*2226

=2223*2219+2223*2225 - 2219*2226 - 2219*2226

=(2223*2219 - 2219*2226) +2223*2225 - 2219*2225 - 2219

=2219*(-3) + 2225*4 - 2219

=2219*(-4) + 2225*4 = 4*(2225-2219) = 4*6 = 24

B=3004*2999*(3001-4) - 3003*2996*3001

=(3003+1)*2999*3001 - 3004*2999*4 - 3003*2996*3001

=3003*(2996+3)*3001 +2999*3001 - 3004*2999*4 - 3003*2996*3001

=3003*2996*3001+3*3003*3001 +2999*3001 - 3*3004*2999 - 3004*2999 - 3003*2996*3001

=3*(3003*3001 - 3004*2999) + 2999*(3001-3004)

=3*[(3004-1)*(2999+2) - 3004*2999] - 3*2999

=3*(3004*2999+2*3004 - 2999 - 2 - 3004*2999) - 3*2999

=3*(2*3003-2999) - 3*2999

=6*3003 - 6*2999 = 6*(3003-2999) = 6*4 = 24

===> A=B (=24)

2003/2004=1 - 1/2004

2002/2003= 1- 2003

Vì: 1/2004< 1/2003 => 1- 1/2004> 1-2003

=> 2003/2004> 2002/2003