a) \(\frac{5}{6}\)= \(\frac{15}{18}\); b)  \(\frac{99}{100}\)< \(\frac{100}{99}\);   c ) \(\frac{15}{17}\)> \(\frac{13}{18}\)vì \(\frac{15}{17}\)> \(\frac{15}{18}\)> \(\frac{13}{18}\); 

d) \(\frac{222}{333}\)= \(\frac{2}{3}\)\(=1-\frac{1}{3}\); \(\frac{3333}{4444}\)= \(\frac{3}{4}\)= \(1-\frac{1}{4}\); vì \(\frac{1}{3}\)> \(\frac{1}{4}\)nên \(\frac{222}{333}\)< \(\frac{3333}{4444}\)

e) \(\frac{292929}{272727}\)= \(\frac{29}{27}\)= \(1+\frac{2}{17}\); \(\frac{347347}{345345}\)= \(\frac{347}{345}\)= \(1+\frac{2}{345}\)nên \(\frac{292929}{272727}\)> \(\frac{347347}{345345}\)

\(\dfrac{333}{337}=\dfrac{337}{337}-\dfrac{4}{337}=1-\dfrac{4}{337}\\ \dfrac{277}{281}=\dfrac{281}{281}-\dfrac{4}{281}=1-\dfrac{4}{281}\\ \)

Ta thấy : \(\dfrac{4}{337}< \dfrac{4}{281}\)

\(=>1-\dfrac{4}{337}>1-\dfrac{4}{281}\\ =>\dfrac{333}{337}>\dfrac{277}{281}\)

ta có: \(M=\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\)

\(M=\frac{7.9+7.2.9.3+7.3.9.4}{21.27+21.227.3+21.3.27.4}\)

\(M=\frac{7.9.\left(1+2.3+3.4\right)}{21.27.\left(1+2.3+3.4\right)}\)

\(M=\frac{7.9}{3.7.3.9}\)

\(M=\frac{1}{3^2}\)

\(M=\frac{1}{9}\)

và \(N=\frac{37}{333}=\frac{37}{37.9}=\frac{1}{9}\)

\(N=\frac{1}{9}\)

\(\Rightarrow M=N\left(=\frac{1}{9}\right)\)

M= \(\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\).

M= \(\frac{7.9+14.27+21.36}{7.3.9.3+14.3.27.3+21.3+36.3}\).

M= \(\frac{7.9+14.27+21.36}{7.9.9+14.27.9+21.36.9}\).

M= \(\frac{7.9+14.27+21.36}{9\left(7.9+14.27+21.36\right)}\).

M= \(\frac{1}{9}\).

Ta có: N= \(\frac{37}{333}\)= \(\frac{1}{9}\).

Vì \(\frac{1}{9}\)= \(\frac{1}{9}\) nên M= N.

Vậy M= N.

giúp mình nha 

-Vì (1/222)^333=(1/222)^3.111=(3/666)^111

     (1/333)^222=(1/333)^2.111=(2/666)^111

-Vì 111=111 và 3/666>2/666

=))(1/222)^333>(1/333)^222

Ta có:

222³³³ = 111³³³.2³³³ = 111²²².111¹¹¹.(2³)¹¹¹ = 111²²².111¹¹¹.8¹¹¹ = 111²²².888¹¹¹

333²²² = 111²²².3²²² = 111²²².(3²)¹¹¹ = 111²²².9¹¹¹

Vì 111²²².888¹¹¹ > 111²²².9¹¹¹

=> 222³³³ > 333²²²

=> \(\frac{1}{222^{333}}< \frac{1}{333^{222}}\)

hay \(\left(\frac{1}{222}\right)^{333}< \left(\frac{1}{333}\right)^{222}\)

a)Ta có: \(\frac{1313}{1515}< \frac{1313}{1428}< \frac{1326}{1428}\Rightarrow\frac{1313}{1515}< \frac{1326}{1428}\)

b)Ta có: \(1-\frac{119}{120}=\frac{1}{120}< 1-\frac{118}{119}=\frac{1}{119}\Rightarrow\frac{119}{120}>\frac{118}{119}\)

c)Ta có: \(\frac{222}{555}< \frac{222}{444}< \frac{333}{444}\Rightarrow\frac{222}{555}< \frac{333}{444}\)

A) <

b) >

c) =

\(\frac{15}{13}\)> 1 

\(\frac{70}{117}\)< 1

=> \(\frac{15}{13}\)> \(\frac{70}{117}\)

k cho mk nha

2 ) So sánh 333^444 và 444^333: 
Có 333^444=(333^4)^111 và 444^333=(444^3)^111 
Như vậy ta cần so sánh 333^4 và 444^3: 
Vì 333^4/444^3=3^4*111^4/(4^3*111^3)=3^4*11... nên 333^4>444^3 do đó 
333^444>444^333 

1,\(\frac{2}{3}x=\frac{3}{4}y=\frac{4}{5}z\Rightarrow\frac{12x}{18}=\frac{12y}{16}=\frac{12z}{15}\)

Aps dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{12x}{18}=\frac{12y}{16}=\frac{12z}{15}=\frac{12x+12y+12z}{18+16+15}=\frac{12\left(x+y+z\right)}{49}=\frac{12.147}{49}=\frac{1764}{49}\)=36

\(\Rightarrow\hept{\begin{cases}x=36.18:12=54\\y=36.16:12=48\\z=36.15:12=45\end{cases}}\)

Vậy:.......

1.

Đặt \(\frac{2}{3}x=\frac{3}{4}y=\frac{4}{5}z=k\)    \(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}k\\y=\frac{4}{3}k\\z=\frac{5}{4}k\end{cases}}\)

mà \(x+y+z=147\) \(\Leftrightarrow\frac{3}{2}k+\frac{4}{3}k+\frac{5}{4}k=147\) \(\Leftrightarrow k\left(\frac{3}{2}+\frac{4}{3}+\frac{5}{4}\right)=147\)

\(\Leftrightarrow\frac{49}{12}k=147\) \(\Leftrightarrow k=147\div\frac{49}{12}=36\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}k=54\\y=\frac{4}{3}k=48\\z=\frac{5}{4}k=45\end{cases}}\)

2.

\(\left(3^4\right)^{111}\times111^{111}>\left(4^3\right)^{111}\)\(\Leftrightarrow3^{444}\times111^{111}\times\left(111^{111}\right)^4>4^{333}\times\left(111^{111}\right)^3\) 

\(\Leftrightarrow3^{444}\times111^{444}>4^{333}\times111^{333}\)

\(\Leftrightarrow333^{444}>444^{333}\)

\(\frac{40}{50}=\frac{4}{5}\)=0,8/1

So sánh 2 phân số 0.8/1 và 10/1

( nếu 2 phan số cùng mẫu thì tử số của phân số nào lớn hơn thì lớn hơn)

=> 0,8/1<10/1

=> 40/50<10/1