= 125 x 1 - 57 = 68 nha e

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

(1/2)^m = 1/32

mà 1/32 = (1/2)^5 nên m = 5

343/125= (7/5)^n

mà 343/125 = (7/5)^3 nên n=3

73 nha~

ƠwƠ

m - m + a : a . 125 - 52 

= 0 + 1 . 125 - 52

=125 - 52

= 73

Sửa đề: \(M=\frac{a}{b}+\frac{b}{a}\)

TA có: \(\frac{a}{b}+\frac{b}{a}\ge2\cdot\sqrt{\frac{a}{b}\cdot\frac{b}{a}}\)

=>M>=2∀a,b<>0

Dấu '=' xảy ra khi a=b

a: N={\(\overline{abc};\overline{acb};\overline{bac};\overline{bca};\overline{cab};\overline{cba}\) }

b: Tổng các phần tử của N là 1998

=>\(\overline{abc}+\overline{acb}+\overline{bac}+\overline{bca}+\overline{cab}+\overline{cba}=1998\)

=>100a+10b+c+100a+10c+b+100b+10a+c+100b+10c+a+100c+10a+b+100c+10b+a=1998

=>222(a+b+c)=1998

=>a+b+c=9

ta ví dụ a/b = 5/4

ta có 5/4 ... 5+1/4+1

      = 5/4 ... 6/5

ta quy đồng được :5/4 = 25/20 ; 6/5 = 24/20

=> a/b > a+m/b+m

Ta có : a/b = a*(b+m)/b*(b+m) = ab+am/b*(b+m)

            a+m/b+m = (a+m)*b/(b+m)*b = ab+bm/b*(b+m)

  Vì  a/b > 1  => a > b     hay am > bm

  Vậy ab+am/b*(b+m) > ab+bm/b*(b+m)   Hay a/b > a+m/b+m