Tìm n ∈ N biết
a)\(\dfrac{625}{5^n}=5\)
b)\(\dfrac{\left(-2\right)^n}{16}=-32\)
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a) \(2\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+...+\dfrac{16}{n\left(n+16\right)}\right)=\dfrac{16}{25}\)
\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{n}-\dfrac{1}{n+16}=\dfrac{8}{25}\)
\(\dfrac{1}{3}-\dfrac{1}{n+16}=\dfrac{8}{25}\)
\(\dfrac{n+13}{3\left(n+16\right)}=\dfrac{8}{25}\)
\(24n+384=25n+325\)
\(25n-24n=384-325\)
\(n=59\)
\(\dfrac{625}{5^n}\)=5
=>\(\dfrac{5^4}{5^n}\) =5
=>\(5^4\) :\(5^n\) = 5
=>\(5^{4-n}\) =\(5^1\)
=>4\(-\)n=1
=>n=4-1
=>n=3
\(\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{27}\right)\)
\(\Rightarrow\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{3}\right)^3\)
\(\Rightarrow n=3\)
\(\left(\dfrac{3}{5}\right)^n=\dfrac{81}{625}\)
\(\Rightarrow\left(\dfrac{3}{5}\right)^n=\left(\dfrac{3}{5}\right)^4\)
\(\Rightarrow n=4\)
a, \(\left(\dfrac{1}{3}\right)^n=\dfrac{1}{27}\Rightarrow\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{3}\right)^3\)
Vì \(\dfrac{1}{3}\ne-1,\dfrac{1}{3}\ne0;\dfrac{1}{3}\ne1\) nên \(n=3\)
Vậy........
b, \(\left(\dfrac{3}{5}\right)^n=\dfrac{81}{625}\Rightarrow\left(\dfrac{3}{5}\right)^n=\left(\dfrac{3}{5}\right)^4\)
Vì \(\dfrac{3}{5}\ne-1,\dfrac{3}{5}\ne0;\dfrac{3}{5}\ne1\) nên \(n=4\)
Vậy..........
Chúc bạn học tốt!!!
\(a. \)
\(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\left(-2\right)^n=-32:4=-8\)
\(\Rightarrow\left(-2\right)^n=\left(-2\right)^3\)
\(\Rightarrow n=3\)
\(b.\)
\(\dfrac{8}{2^n}=2\)
\(\Rightarrow2^n=4\)
\(\Rightarrow2^n=2^2\)
\(\Rightarrow n=2\)
\(c.\)
\(\dfrac{16}{\left(-2\right)^n}=-8\)
\(\Rightarrow\left(-2\right)^n=-2\)
\(\Rightarrow n=1\)
a)\(\left(\dfrac{1}{2}\right)^n=\dfrac{1}{32}\)
=>\(\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{2}\right)^5\)
=>n=5
b)\(\left(\dfrac{343}{125}\right)=\left(\dfrac{7}{5}\right)^n\)
=>\(\left(\dfrac{7}{5}\right)^3=\left(\dfrac{7}{5}\right)^n\)
=>n=3
c)\(\dfrac{16}{2^n}=2\)
=>2n=\(\dfrac{16}{2}\)
=>2n=8
=>2n=23
=>n=3
d)\(\dfrac{\left(-3\right)^n}{81}=-27\)
=>(-3)n=-27.81
=>(-3)n=-2187
=>(-3)n=(-3)7
=>n=7
e)8n:2n=4
=>(23)n:2n=4
=>23n:2n=4
=>23n-n=4
=>22n=4
=>22n=22
=>2n=2
=>n=1
f)32.3n=35
=>3n=35:32
=>3n=35-2
=>3n=33
=>n=3
g) (22:4).2n=4
=>1.2n=22
=>n=2
h)3-2.34.3n=37
=>\(\left(\dfrac{1}{3}\right)^2\).34.3n=37
=>32.3n=37
=>32+n=37
=>2+n=7
=>n=5
c)\(7^{2n}+7^{2n+2}=2450\)
⇒\(7^{2n}+7^{2n}.7^2=2450\)
⇒\(7^{2n}.50=2450\)
⇒\(7^{2n}=49\)\(=7^2\)
⇒2n=2
⇒n=1
a)\(\dfrac{625}{5^n}=5\)\(\Rightarrow\)\(\dfrac{5^4}{5^n}=5\)\(\Rightarrow\)\(5^n=5^5:5^4=5^1\Rightarrow n=1\)
b)\(\dfrac{\left(-2\right)^n}{16}=-32\Rightarrow\)\(\left(-2\right)^n=-32.16=\left(-1\right).2^5.2^4=\left(-1\right).2^9=\left(-2\right)^9\)
.............................\(\Rightarrow n=9\)
a. \(\dfrac{625}{5^n}=5\Rightarrow5^n=\dfrac{625}{5}=125\Rightarrow5^n=5^3\Rightarrow n=3\left(TMĐK\right)\)
Vậy n=3
b. \(\dfrac{\left(-2\right)^n}{16}=-32\Rightarrow\left(-2\right)^n=-32\cdot16=-512\Rightarrow\left(-2\right)^n=\left(-2\right)^9\Rightarrow n=9\left(TMĐK\right)\)
Vậy n=9