Tìm các số tự nhiên m và n biết:
a)\(\left(0,125\right)^m.24^m=243\)
b)\(85^m:17^n=625\)
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a) Ta có: \(\left(0.125\right)^m\cdot24^m=243\)
\(\Leftrightarrow3^m=3^5\)
hay m=5
a) Ta có : (x - 4)2 = (x - 4)4
=> (x - 4)4 - (x - 4)2 = 0
=> (x - 4)2.[(x - 2)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-4\right)^2=0\\\left(x-4\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-4=0\\\left(x-4\right)^2=1\end{cases}}\)
Nếu x - 4 = 0 => x = 4
Nếu (x - 4)2 = 1 => \(\orbr{\begin{cases}x-4=1\\x-4=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}\)
Vậy \(x\in\left\{3;4;5\right\}\)
2) a. Bạn xem lại đề
b) Ta có : 85n : 17n= 625
=> (85 :17)n = 625
=> 5n = 625
=> 5n = 53
=> n = 3
Vậy n = 3
\(\left(\dfrac{1}{8}\right)^m\cdot24^m=243\)
\(\Rightarrow\left(\dfrac{1}{8}\cdot24\right)^m=243\)
\(\Rightarrow3^m=243=3^5\)
\(\Rightarrow m=5\)
(0,125)m . 24m = 243
(0,125 . 24)m =243
3m =35
=>m = 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) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(=>\left(\frac{1}{2}\right)^m=\frac{1^5}{2^5}\)
\(=>\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
\(=>m=5\)
b) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(=>\frac{7^3}{5^3}=\left(\frac{7}{5}\right)^n\)
\(=>\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
\(=>n=3\)
a) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(\Rightarrow\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
=> m =5
b) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(\Rightarrow\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
=> n = 3
a) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(\Rightarrow\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
=> m = 5
Vậy m = 5
b) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(\Rightarrow\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
=> n = 3
Vậy n = 3
a) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}=\left(\frac{1}{2}\right)^5\)
=> m= 5
b) \(\left(\frac{7}{5}\right)^n=\frac{243}{125}=\left(\frac{7}{5}\right)^3\)
=> n = 3