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a) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\frac{1^4}{3^4}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow n=4\)
Vậy n = 4
b) \(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\frac{-8^3}{7^3}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\left(\frac{-8}{7}\right)^3=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow n=3\)
Vậy n = 3
a)
\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)
Vậy \(x = \frac{{ - 2}}{3}\).
b)
\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)
Vậy\(x = \frac{1}{12}\).
c)
\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)
Vậy \(x = \frac{7}{3}\).
d)
\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)
Vậy \(x = \frac{{ - 9}}{{10}}\).
a)
\(\begin{array}{l}x + \frac{1}{2} = - \frac{1}{3}\\x = - \frac{1}{3} - \frac{1}{2}\\x = - \frac{2}{6} - \frac{3}{6}\\x = \frac{{ - 5}}{6}\end{array}\)
Vậy \(x = \frac{{ - 5}}{6}\).
b)
\(\begin{array}{l}\left( { - \frac{2}{7}} \right) + x = - \frac{1}{4}\\x = - \frac{1}{4} - \left( { - \frac{2}{7}} \right)\\x = - \frac{1}{4} + \frac{2}{7}\\x = - \frac{7}{{28}} + \frac{8}{{28}}\\x = \frac{1}{{28}}\end{array}\)
Vậy \(x = \frac{1}{{28}}\).
\(\left(\frac{-3}{4}\right)^{3x-1}=\frac{256}{81}\)
\(\Rightarrow\left(\frac{-3}{4}\right)^{3x}:\left(\frac{-3}{4}\right)=\frac{256}{81}\)
\(\Rightarrow\left(\frac{-3}{4}\right)^{3x}=\frac{256}{81}.\left(\frac{-3}{4}\right)\)
\(\Rightarrow\left(\frac{-3}{4}\right)^{3x}=\frac{64}{27}\)
\(\Rightarrow\left(-\frac{3}{4}\right)^{3x}=\left(-\frac{4}{3}\right)^3\)
\(\Rightarrow1=\left(-\frac{4}{3}\right)^3.\left(-\frac{4}{3}\right)^{3x}\)
\(\Rightarrow1=\left(-\frac{4}{3}\right)^{-x}\)
\(\Rightarrow1=\left(-\frac{3x}{4}\right)\)
\(\Rightarrow1=-3x:4\)
\(\Rightarrow-3x=4\)
\(\Rightarrow x=-\frac{4}{3}\)
Ta có:
\(\frac{256}{81}=\frac{4^4}{"-3"^4}=\frac{1}{\frac{"-3"^4}{4}}=\frac{1}{"\frac{-3}{4}"^4}="\frac{-3}{4}"^4="\frac{-3}{4}"^{3x-1}\Rightarrow3x-1=-4\Rightarrow3x=-4+1\)
\(=-3\)
\(\Rightarrow x=-3:1=-1\)
\(a.172x^2-7^9=2^{-3}.98^3=117649\)
\(172x^2=117649+7^9=40471256\)
\(x^2=40471256:172=235298\)
\(x=\sqrt{235298}=485.07......\)
a)
\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} = - \frac{1}{2}\\x = - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
b)
\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)
Vậy \(x = \frac{9}{{25}}\).
c)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)
Vậy \(x = \frac{4}{9}\).
d)
\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
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!
b) \(\left(-\frac{3}{4}\right)^{3m-1}=\left(\frac{4}{3}\right)^4=\left(\frac{1}{\frac{3}{4}}\right)^4=\left(\left(\frac{3}{4}\right)^{-1}\right)^4=\left(\frac{3}{4}\right)^{-4}=\left(-\frac{3}{4}\right)^{-4}\) (Lũy thừa số mũ chẵn thì am = (-a)m)
=> 3m - 1 = -4 => 3m = -3 => m = -1
a) \(7^{m-1}=\frac{343}{345}\) => không tồn tại số nguyên m thỏa mãn
a) \(7^{m+2}+2.7^{m-1}=343\)
\(7^{m-1}.7^3+2.7^{m-1}=343\)
\(7^{m-1}.\left(7^3+2\right)=343\)
\(7^{m-1}.345=343\)
\(7^{m-1}=\frac{343}{345}\)
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