Tìm n, biết:
a) \(\dfrac{-32}{\left(-2\right)^n}=4\)
b) \(\dfrac{8}{2^n}=2\)
c) \(\dfrac{16}{\left(-2\right)^n}=-8\)
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a) \(\dfrac{32}{\left(-2\right)^n}=4\)
\(\Rightarrow\left(-2\right)^n=8=\left(-2\right)^3\)
=> n = 3
b) \(\dfrac{8}{2^n}=2\)
\(\Rightarrow2^n=4=2^2\)
=> n = 2
c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
=> 2n - 1 = 3
=> 2n = 4
=> n = 2
\(\left(-2\right)^3=-8\) bạn ạ chứ không phải là bằng 8 nên n = 3 là không đúng rồi
Bài 1:
a, \(\left(x-2\right)^2=9\)
\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)
b, \(\left(3x-1\right)^3=-8\)
\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)
\(\Rightarrow x=-\dfrac{1}{3}\)
c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)
\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)
d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)
Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)
e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)
Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)
f, \(\left(\dfrac{1}{2}\right)^{2x-1}=8\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^{-3}\) Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(2x-1=-3\) \(\Rightarrow2x=-2\Rightarrow x=-1\) Chúc bạn học tốt!!!a. \(\dfrac{32}{2^n}=2\)
\(\Leftrightarrow2^n.2=32\)
\(\Rightarrow2^{n+1}=2^5\)
\(\Rightarrow n+1=5\)
\(\Rightarrow n=4\)
Vậy...
b. \(16^n:2^n=8\)
\(\Rightarrow\left(2^4\right)^n:2^n=2^3\)
\(\Rightarrow4n-n=3\)
\(\Rightarrow3n=3\)
\(\Rightarrow n=1\)
Vậy...
c. \(\dfrac{\left(-3\right)^n}{81}=\left(-27\right)\)
\(\Leftrightarrow\left(-3\right)^n=81.\left(-27\right)\)
\(\Rightarrow\left(-3\right)^n=\left(-2187\right)\)
\(\Rightarrow\left(-3\right)^n=\left(-3\right)^7\)
\(\Rightarrow n=7\)
Vậy...
a, \(\dfrac{32}{2^n}\) =2 =>25=2n+1=>5=n+1=>n=4
b, 24n:2n=23=>23n=23=>3n=3=>n=1
c, \(\dfrac{\left(-3\right)^n}{81}\)=(-27)=>(-3)n=34 . (-3)3=>n=7
a) \(2\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+...+\dfrac{16}{n\left(n+16\right)}\right)=\dfrac{16}{25}\)
\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{n}-\dfrac{1}{n+16}=\dfrac{8}{25}\)
\(\dfrac{1}{3}-\dfrac{1}{n+16}=\dfrac{8}{25}\)
\(\dfrac{n+13}{3\left(n+16\right)}=\dfrac{8}{25}\)
\(24n+384=25n+325\)
\(25n-24n=384-325\)
\(n=59\)
1. Tìm n, biết:
a) \(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
\(\Rightarrow\left(-2\right)^n.\left(-2\right)^2=\left(-2\right)^5\)
(-2)n + 2 = (-2)5
n + 2 = 5
n = 5 - 2
n = 3.
b) \(\dfrac{8}{2^n}=2\)
\(\Rightarrow\dfrac{2^3}{2^n}=2\)
\(\Rightarrow\) 2n . 2 = 23
n + 1 = 3
n = 3 - 1
n = 2.
c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
2n - 1 = 3
2n = 3 + 1
2n = 4
n = 4 : 2
n = 2.
2. Tính:
a) \(\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left[\left(\dfrac{1}{2}\right)^2\right]^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{2}\right)^4\)
\(=\left(\dfrac{1}{2}\right)^7\)
\(=\dfrac{1}{128}\)
b) 273 : 93
= (33)3 : (32)3
= 39 : 36
= 33
= 27
c) 1252 : 253
= (53)2 : (52)3
= 56 : 56
= 1
d) \(\dfrac{27^2.8^5}{6^6.32^3}\)
\(=\dfrac{\left(3^3\right)^2.\left(2^3\right)^5}{6^6.\left(2^5\right)^3}\)
\(=\dfrac{3^6.2^{15}}{6^6.2^{15}}\)
\(=\dfrac{3^6}{6^6}\)
\(=\dfrac{1}{64}.\)
B2 :
b) 27\(^3\): 9\(^3\)= (27:9)\(^3\)= 3\(^3\)
c) 125\(^2\): 25\(^3\)= 15625 : 15625 = 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
\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)
\(a. \)
\(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\left(-2\right)^n=-32:4=-8\)
\(\Rightarrow\left(-2\right)^n=\left(-2\right)^3\)
\(\Rightarrow n=3\)
\(b.\)
\(\dfrac{8}{2^n}=2\)
\(\Rightarrow2^n=4\)
\(\Rightarrow2^n=2^2\)
\(\Rightarrow n=2\)
\(c.\)
\(\dfrac{16}{\left(-2\right)^n}=-8\)
\(\Rightarrow\left(-2\right)^n=-2\)
\(\Rightarrow n=1\)