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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) (12)m=132(12)m=132
\(\Rightarrow\left(\dfrac{1}{2}\right)^m=\left(\dfrac{1}{2}\right)^5\Rightarrow m=5\)
b)
343125=(75)n
\(\Rightarrow\left(\dfrac{7}{5}\right)^3=\left(\dfrac{7}{5}\right)^n\Rightarrow n=3\)

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) \(\left(2^2:4\right).2^n=2=32\)
\(\Leftrightarrow\left(2^2:2^2\right).2^n=2^5\)
\(\Leftrightarrow2^{2-2}.2^n=2^5\)
\(\Leftrightarrow2^0.2^n=2^5\)
\(\Leftrightarrow1.2^n=2^5\)
\(\Leftrightarrow2^n=2^5\)
\(\Leftrightarrow n=5\)
Vậy n=5
b) \(27< 3^n< 243\)
\(\Leftrightarrow3^3< 3^n< 3^5\)
\(\Leftrightarrow3< n< 5\)
\(\Rightarrow n=4\)
Vậy n=4
a. \(\left(2^2:4\right).2^n=32\)
\(\Rightarrow\left(4:4\right).2^n=32\)
\(\Rightarrow2.2^n=32\)
\(\Rightarrow2^n=32:1=32=2^5\)
\(\Rightarrow n=5\)
Vậy................
b. \(27< 3^n< 243\)
\(\Leftrightarrow3^3< 3^n< 3^5\)
\(\Rightarrow3< n< 5\)
\(\Rightarrow n=4\)
c. đề bài có j đó sai sai
Kẻ DB//AM => BAM+ABD=1800(2 góc trong cùng phía)
hay 1250+ABD=1800
=>ABD=1800-1250=550
Ta có: BD//CN =>BDC+BCN=1800(2 góc trong cùng phía)
hay 1400+BCN =1800
=>BCN=1800-1400=400
=>ABC=ABD+BCN=400+400=800
NHỚ KÍ HIỆU GÓC
Kẻ đường thẳng Bp song song với m và n
ta có : \(\hept{\begin{cases}\widehat{pBA}=180^0-A=55^0\\\widehat{pBC}=180^0-C=40^0\end{cases}\Rightarrow B=}\widehat{pBA}+\widehat{pBC}=95^0\)