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\(\)Mk ko ghi lại đề đâu nha
\(Xet2TH:\left(+\right)n\ge2018\Rightarrow|n-2018|=n-2018\Rightarrow2018^m+4035=2n-2018\)
\(2n-2018\left(chẵn\right)\Rightarrow2018^mlẻ\Rightarrow m=0\Rightarrow2n-2018=4036\Rightarrow n=3027\)
\(\left(+\right)n< 2018\Rightarrow|n-2018|=2018-n\Rightarrow2018^m+4035=2018.Mà:2018^m\ge0\left(loại\right)\)
\(Vậy:m=0;n=3027\)
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) \(\left(1:2\right)^m=1:32\Leftrightarrow\left(\frac{1}{2}\right)^m=\frac{1}{32}\Leftrightarrow\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\Rightarrow m=5\)
b) \(343:125=\left(7:5\right)^n\Leftrightarrow\frac{343}{125}=\left(\frac{7}{5}\right)^n\Leftrightarrow\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\Rightarrow n=3\)
a, \(\left(1:2\right)^m=1:32=\left(1:2\right)^5\Rightarrow m=5\)
b, \(343:125=\left(7:5\right)^n\Rightarrow\left(7:5\right)^3=\left(7:5\right)^n\Rightarrow n=3\)
a, ( 1/2 ) ^ m = ( 1/2) ^5
=> m = 5
b, ( 7/5) ^n = 343 / 125
=> ( 7/5)^n = (7/5) ^ 3
=> n = 3
Đúng cho tui nha
\(a.\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^m=\frac{1^5}{2^5}\)
\(\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
=>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
k em nha em mới lớp 5