tìm m \(\in\) Z, biết:
a.7m+2 + 2.7m-1 = 343
b.\(\left(-\frac{3}{4}\right)^{3m-1}=\frac{256}{81}\)
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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:
\(\frac{256}{81}=\frac{4^4}{"-3"^4}=\frac{1}{\frac{"-3"^4}{4}}=\frac{1}{"\frac{-3}{4}"^4}="\frac{-3}{4}"^4="\frac{-3}{4}"^{3x-1}\Rightarrow3x-1=-4\Rightarrow3x=-4+1\)
\(=-3\)
\(\Rightarrow x=-3:1=-1\)
\(a.172x^2-7^9=2^{-3}.98^3=117649\)
\(172x^2=117649+7^9=40471256\)
\(x^2=40471256:172=235298\)
\(x=\sqrt{235298}=485.07......\)
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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