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\(27^n:3^n=\left(27:3\right)^n=9\)
\(9^n=9\rightarrow n=1\)
\(\left(\frac{25}{5}\right)^n=5^n=5^1\)
\(\rightarrow n=1\)
\(\frac{81}{\left(-3\right)^n}=-243=\left(-3\right)^5\)
\(\rightarrow\left(-3\right)^n=81:\left(-3\right)^5=\frac{-1}{3}=\left(-3\right)^{-1}\)
\(\)
a: =>9^n=9
=>n=1
b: =>5^n=5
=>n=1
c: \(\Leftrightarrow\left(-27\right)^n=-243\)
=>\(\left(-3\right)^{3n}=\left(-3\right)^5\)
=>3n=5
=>n=5/3
d: =>2^n*9/2=9*2^5
=>2^n=9*2^5:9/2=2^5*2=2^6
=>n=6
\(\text{Bn hỏi từ từ từng câu 1 thôi}\)
\(\text{Bn hỏi thế ai mà dám làm}\)
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Chí lí
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sọ ghi 2 hàng khoogn đc tích tăng lê hiều hàng
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c, \(\frac{-32}{-2^n}=4\)
\(\Rightarrow-2^n=-32:4\)
\(\Rightarrow-2^n=-8\)
\(\Rightarrow-2^n=-2^3\Rightarrow n=3\)
d, \(\frac{8}{2^n}=2\)
\(\Rightarrow2^n=8:2\)
\(\Rightarrow2^n=4\)
\(\Rightarrow2^n=2^2\Rightarrow n=2\)
e, \(\frac{25^3}{5^n}=25\)
\(\Rightarrow5^n=25^3:25\)
\(\Rightarrow5^n=25^2\)
\(\Rightarrow5^n=5^4\Rightarrow n=4\)
i , \(8^{10}:2^n=4^5\)
\(\Rightarrow2^n=8^{10}:4^5\)
\(\Rightarrow2^n=\left(2^3\right)^{10}:\left(2^2\right)^5\)
\(\Rightarrow2^n=2^{30}:2^{10}\)
\(\Rightarrow2^n=2^{20}\Rightarrow n=20\)
k, \(2^n.81^4=27^{10}\)
\(\Rightarrow2^n=27^{10}:81^4\)
\(\Rightarrow2^n=\left(3^3\right)^{10}:\left(3^4\right)^4\)
\(\Rightarrow2^n=3^{30}:3^{16}\)
\(\Rightarrow2^n=3^{14}\)
\(\Rightarrow2^n=4782969\)Không chia hết cho 2 nên ko có Gt n thỏa mãn
\(\frac{1}{2}.2^n+4.2^n=9.2^5\Rightarrow2^n\left(\frac{1}{2}+4\right)=288\Rightarrow2^n.\frac{9}{2}=288\Rightarrow2^{n-2}.9=288\Rightarrow2^{n-2}=32\)(dấu "=>" số 3 bn sửa thành 2n-1.9=288=>2n-1=32 nha)
=>2n-1=25=>n-1=5=>n=5+1=6
vậy......
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b,\(D=2.\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{n.\left(n+2\right)}\right)\)
\(\Rightarrow D=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{n.\left(n+2\right)}\)
\(\Rightarrow D=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n.\left(n+2\right)}\)
\(\Rightarrow D=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+2}\)
\(\Rightarrow D=1-\frac{1}{n+2}=\frac{n}{n+2}< \frac{n+2}{n+2}=1\left(1\right)\)
\(\Rightarrow D=\frac{n}{n+2}>0\left(2\right)\)
Từ (1);(2)\(\Rightarrow0< D< 1\)
\(\Rightarrowđpcm\)
a,\(C>0\)
\(C=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{19}< 9;\frac{1}{11}< 1\)
\(\Rightarrow0< A< 1\)
\(\Rightarrow A\notinℤ\)
c,\(E=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}\)
Ta quy đồng 3 số đầu
\(=\frac{2}{6}+\frac{2}{8}+\frac{2}{10}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}>\frac{6.2}{12}=1\)
\(E=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}\)
\(=\frac{2}{6}+\frac{2}{8}+\frac{2}{10}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}< \frac{6.2}{6}=2\)
\(1< E< 2\)
\(E\notinℤ\)
a) \(\dfrac{81}{\left(-3\right)^n}=-243\)
\(\dfrac{\left(-3\right)^4}{\left(-3\right)^n}=\left(-3\right)^5\)
\(\left(-3\right)^n=\dfrac{\left(-3\right)^4}{\left(-3\right)^5}=\left(-3\right)^{-1}\)
n = -1
Vậy n = -1
b) \(\dfrac{25}{5^n}=5\)
\(\dfrac{5^2}{5^n}=5^1\)
\(5^n=\dfrac{5^2}{5^1}=5^1\)
n = 1
Vậy n = 1
c) \(\dfrac{1}{2}\cdot2^n+4\cdot2^n=9\cdot2^5\)
\(2^{n-1}+4\cdot2^{n-1}\cdot2=9\cdot2^5\)
\(2^{n-1}+8\cdot2^{n-1}=9\cdot2^5\)
\(\left(8+1\right)\cdot2^{n-1}=9\cdot2^5\)
\(9\cdot2^{n-1}=9\cdot2^5\)
\(2^{n-1}=2^5\cdot\dfrac{9}{9}=2^5\)
n - 1 = 5
n = 5 + 1 = 6
Vậy n = 6
a) 81/(-3)ⁿ = -243
(-3)ⁿ = 81 : (-243)
(-3)ⁿ = -1/3
n = -1
b) 25/5ⁿ = 5
5ⁿ = 25 : 5
5ⁿ = 5
n = 1
c) 1/2 . 2ⁿ + 4 . 2ⁿ = 9 . 2⁵
2ⁿ . (1/2 + 4) = 9 . 32
2ⁿ . 9/2 = 288
2ⁿ = 288 : 9/2
2ⁿ = 64
2ⁿ = 2⁶
n = 6