a) 2711= 32×11=322; 818=34×8=332
=>2711<818
a. \(27^{11}\) và \(81^8\)
\(27^{11}=\left(3^3\right)^{11}=3^{33}\)
\(81^8=\left(3^4\right)^8=3^{32}\)
Mà \(3^{33}>3^{32}\Rightarrow27^{11}>81^8\)
b. \(625^5\) và \(125^7\)
\(625^5=\left(5^4\right)^5=5^{20}\)
\(125^7=\left(5^3\right)^7=5^{21}\)
Mà \(5^{20}< 5^{21}\Rightarrow626^5< 125^7\)
c. \(3^{2.n}\) và \(2^{3.n}\)
\(3^{n.2}=9n\)
\(2^{3.n}=8n\)
Mà \(9n>8n\Rightarrow3^{2.n}>2^{3.n}\)
d. \(29^{18}>79^{13}\)
e. \(3^{39}\) và \(11^{21}\)
\(3^{39}< 3^{40}=\left(3^2\right)^{20}=9^{20}< 11^{21}\)
\(\Rightarrow3^{39}< 11^{21}\)
a)\(27^{11}\)và \(81^8\)
Ta có : \(27^{11}\)=(\(3^3\))\(^{11}\)=\(3^{33}\)
\(81^8\)=(\(3^4\))\(^8\)=\(3^{32}\)
Vì \(3^{33}\)> \(3^{32}\)nên \(27^{11}\)>\(81^8\)
b)\(625^5\)và \(125^7\)
Ta có: \(625^5\)=(\(5^4\))\(^5\)=\(5^{20}\)
\(125^7\)=(\(5^3\))\(^7\)=\(5^{21}\)
Vì \(5^{20}\)<\(5^{21}\)nên \(625^5\)<\(125^7\)
c)\(3^{2.n}\)và \(2^{3.n}\)
Ta thấy : 2.n và 3.n đều có chung n nên:
ta chỉ so sánh 3^2 và 2^3 . ta có :
\(3^{2.n}\)=\(3^2\)=9
\(2^{3.n}\)=\(2^3\)=8
Vì 9>8 nên \(3^{2.n}\)>\(2^{3.n}\)
d)
2. tìm x
a. \(2^x.4=128\)
\(2^x=128:4=32=2^5\)
\(\Rightarrow x=5\)
Vậy...
b. \(\left(2x-1\right)^3=125\)
\(\left(2x-1\right)^3=5^3\)
\(\Rightarrow2x-1=5\)
\(\Rightarrow2x=5+1=6\)
\(\Rightarrow x=6:2=3\)
Vậy x = 3
c. \(\left(4x+1\right)^5-125=3002\)
\(\left(4x+1\right)^5=3002+123=3125=5^5\)
\(\Rightarrow4x+1=5\)
\(\Rightarrow4x=5-1=4\)
\(\Rightarrow x=4:4=1\)
Vậy x = 1
d. \(\left(3x-2\right)^4+29=227\) => sai đề
\(\left(3x-2\right)^4-29=227\)
\(\left(3x-2\right)^4=227+29=256=4^4\)
\(\Rightarrow3x-2=4\)
\(\Rightarrow3x=4+2=6\)
\(\Rightarrow x=6:3=2\)
Vậy x = 2
d. \(3^x.3.3^2=243\)
\(3^x.3.3^2=3^5\)
\(\Rightarrow3^{x+1+2}=3^5\)
\(\Rightarrow x+1+2=5\)
\(\Rightarrow x=5-1-2=2\)
Vậy x = 2
b) 32, 33, 34 , 35.