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a)
1/9 . 34.3n=37
=>3-2.34.3n=37
=>3-2+4+n=37
=>-2+4+n=7
=>n=7-(-2)-4
=>n=5
#)Giải :
\(\frac{1}{9}.3^4.3^n=3^7\)
\(\frac{1}{9}.81.3^n=3^7\)
\(9.3^n=3^7\)
\(3^2.3^n=3^7\)
\(\Rightarrow2+n=7\)
\(\Rightarrow n=5\)
#~Will~be~Pens~#
a Ta có
1/9.3^4.3^n=3^7
=> 1/3^2.3^4.3^n=3^7
=> 3^2.3^n=3^7
=>3^2+n=3^7
=> 2+n=7
=> n=5( tick nhé)
1) \(32< 2^n< 128\)
\(\Rightarrow2^5< 2^n< 2^7\)
Vì \(5< n< 7\)
Nên \(n=6\)
Vậy \(32< 2^6< 128\)
2) \(2.16\ge2^n>4\)
\(\Rightarrow2^5\ge2^n>2^2\)
Vì \(5\ge n>4\)
nên \(n=5\)
Vậy \(2.16\ge2^5>4\)
3/ Tương tự
P/S: chỉ cần đổi các số ra lũy thừa là sẽ tính được!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Kết bạn với mình nha!
a)16^n<128^4
=>(2^4)^n<(2^7)^4
=>2^4n<2^28
=>4n<28
=>n<28:4
=>n<7
=>n E {0;1;2;3;4;5;6}
b)32<2^n<128
=>2^5<2^n<2^7
=>5<n<7
=>n=6
c)2.16>2^n>4
=>2.2^4>2^n>2
=>2<2^n<2^5
=>1<n<5
=>n E {2;3;4}
tick nhé
đó giúp mk đi mà
à, mk quên chưa nói là ai giúp mk sẽ được luôn 2SP đó
giúp mk nha
cảm ơn nhiều!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\(a,2.16\le2^n< 4\)
\(\Rightarrow2^5\le2^n< 2^2\)
\(\Rightarrow n\in5,4,3\)
\(b,3^2.3^n=3^5\)
\(\Rightarrow3^n=3^5:3^2\)
\(\Rightarrow3^n=3^2\)
\(\Rightarrow n=2\)
\(c,\left(2^2:4\right):2^n=4\)
\(\Rightarrow1:2^n=4\)
\(\Rightarrow2^n=1:4=\frac{1}{4}\)
mà \(n\in N\)nên n ko có giá trị nào
\(d,\frac{1}{9}.3^4.3^n=3^7\)
\(\Rightarrow\left(\frac{1}{3}\right)^2.3^2.3^2.3^n=3^7\)
\(\Rightarrow3^2.3^n=3^7\)
\(\Rightarrow3^n=3^7:3^2\)
\(\Rightarrow3^n=3^5\)
\(\Rightarrow n=5\)
e,f làm tự làm tiếp nhé,mỏi tay wa
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
\(a,32< 2^n< 128\)
\(=>2^5< 2^n< 2^7\)
\(=>n=6\)
Vậy...
\(b,2.16\ge2^n>4\)
\(=>2^5\ge2^n>2^2\)
\(=>n\in\left\{3;4;5\right\}\)
Vậy...
\(c,3^2.3^n=3^5\)
\(3^n=3^5:3^2\)
\(3^n=3^3\)
\(=>n=3\)
Vậy...
\(d,\left(2^2:4\right).2^n=4\)
\(\left(2^2:2^2\right).2^n=4\)
\(1.2^n=4\)
\(2^n=4:1\)
\(2^n=4\)
\(=>2^n=2^2\)
\(=>n=2\)
Vậy ...
\(e,\dfrac{1}{9}.3^4.3^n=3^7\)
\(\dfrac{1}{9}.81.3^n=3^7\)
\(3^2.3^n=3^7\)
\(3^n=3^7:3^2\)
\(3^n=3^5\)
\(=>n=5\)
Vậy...
\(g,\dfrac{1}{2}.2^n+4.2^n=9.2^5\)
\(\left(\dfrac{1}{2}+4\right).2^n=9.2^5\)
\(\dfrac{9}{2}.2^n=9.32\)
\(\dfrac{9}{2}.2^n=288\)
\(2^n=288:\dfrac{9}{2}\)
\(2^n=2^6\)
\(=>n=6\)
Vậy...
a) \(32< 2^n< 128\\ \Rightarrow2^5< 2^n< 2^7\\ \Rightarrow5< n< 7\)
Mà: \(n\inℕ^∗\)
\(\Rightarrow n=6\)
b) \(2.16\ge2^n>4\\ \Rightarrow2^1.2^4\ge2^n>2^2\\ \Rightarrow2^5\ge2^n>2^2\\ \Rightarrow5\ge n>2\)
Mà: \(n\inℕ^∗\)
\(\Rightarrow n\in\left\{5;4;3\right\}\)
c) \(3^2.3^n=3^5\\ \Rightarrow3^{n+2}=3^5\\ \Rightarrow n+2=5\\ \Rightarrow n=3\left(nhận\right)\)