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ta có : \(3^2.\frac{1}{243}.81^2.\frac{1}{3^2}=\frac{3^2.\left(3^4\right)^2}{3^5.3^2}=\frac{3^2.3^8}{3^5.3^1}=3^3\)
a) Câu này thiếu đề nhé bạn.
b) \(\frac{25}{5^n}=5\)
\(\Rightarrow5^n=25:5\)
\(\Rightarrow5^n=5\)
\(\Rightarrow5^n=5^1\)
\(\Rightarrow n=1\)
Vậy \(n=1.\)
c) \(\frac{81}{\left(-3\right)^n}=-243\)
\(\Rightarrow\left(-3\right)^n=81:\left(-243\right)\)
\(\Rightarrow\left(-3\right)^n=-\frac{1}{3}\)
\(\Rightarrow\left(-3\right)^n=\left(-3\right)^{-1}\)
\(\Rightarrow n=-1\)
Vậy \(n=-1.\)
e) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow n=4\)
Vậy \(n=4.\)
f) \(\left(-\frac{3}{4}\right)^n=\frac{81}{256}\)
\(\Rightarrow\left(-\frac{3}{4}\right)^n=\left(-\frac{3}{4}\right)^4\)
\(\Rightarrow n=4\)
Vậy \(n=4.\)
Chúc bạn học tốt!
d) \(\frac{1}{2}.2^n+4.2^n=9.2^5\)
\(\Rightarrow2^n.\left(\frac{1}{2}+4\right)=288\)
\(\Rightarrow2^n.\frac{9}{2}=288\)
\(\Rightarrow2^n=288:\frac{9}{2}\)
\(\Rightarrow2^n=64\)
\(\Rightarrow2^n=2^6\)
\(\Rightarrow n=6\)
Vậy \(n=6.\)
g) \(-\frac{512}{343}=\left(-\frac{8}{7}\right)^n\)
\(\Rightarrow\left(-\frac{8}{7}\right)^n=\left(-\frac{8}{7}\right)^3\)
\(\Rightarrow n=3\)
Vậy \(n=3.\)
h) \(5^{-1}.25^n=125\)
\(\Rightarrow5^{-1}.5^{2n}=5^3\)
\(\Rightarrow5^{-1+2n}=5^3\)
\(\Rightarrow-1+2n=3\)
\(\Rightarrow2n=3+1\)
\(\Rightarrow2n=4\)
\(\Rightarrow n=4:2\)
\(\Rightarrow n=2\)
Vậy \(n=2.\)
k) \(3^{-1}.3^n+6.3^{n-1}=7.3^6\)
\(\Rightarrow3^{n-1}+6.3^{n-1}=7.3^6\)
\(\Rightarrow3^{n-1}.\left(1+6\right)=7.3^6\)
\(\Rightarrow3^{n-1}.7=7.3^6\)
\(\Rightarrow n-1=6\)
\(\Rightarrow n=6+1\)
\(\Rightarrow n=7\)
Vậy \(n=7.\)
Chúc bạn học tốt!
a: \(=3^2\cdot3^5:3^4=3^{2+5-4}=3^3\)
b: \(=2^3\cdot2^4:\left(\dfrac{8}{16}\right)=\dfrac{2^7}{2}=2^6\)
c: \(=3^7\cdot3^3=3^{10}\)
d: \(=5^3\cdot5^2\cdot\dfrac{1}{5^4}=5^1\)
Bài 3:
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\Rightarrow x-2=\pm1\)
+) \(x-2=1\Rightarrow x=3\)
+) \(x-2=-1\Rightarrow x=1\)
Vậy \(x=3\) hoặc \(x=1\)
c) \(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=\frac{-1}{2}\)
Vạy \(x=\frac{-1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(\Rightarrow x=\frac{-1}{4}\)
Vậy \(x=\frac{-1}{4}\)
a) \(\left(\frac{27}{64}\right)^8:\left(\frac{3}{4}\right)^{22}=\left[\left(\frac{3}{4}\right)^3\right]^8:\left(\frac{3}{4}\right)^{22}=\left(\frac{3}{4}\right)^{24}:\left(\frac{3}{4}\right)^{22}=\left(\frac{3}{4}\right)^2\)
b) \(\left(\frac{2}{3}\right)^2.\left(-\frac{8}{27}\right).\left(-\frac{2}{3}\right)=\left(\frac{2}{3}\right)^2.\left(-\frac{2}{3}\right)^3.\left(-\frac{2}{3}\right)=\left(\frac{2}{3}\right)^2.\left(-\frac{2}{3}\right)^4=\left(\frac{2}{3}\right)^6\)