a) \(A=\frac{135}{135.136-1}\)             và                    \(B=\frac{136}{136.137-1}\)

   \(A=\frac{1}{136-1}=\frac{1}{135}\)                             \(B=\frac{1}{137-1}=\frac{1}{136}\)

Vì \(\frac{1}{136}\)< \(\frac{1}{135}\)nên A > B.

a, A = \(\frac{136-1}{\left(136-1\right)136-1}\) = \(\frac{136-1}{136^2-136-1}\)                 B=\(\frac{136}{136\left(136+1\right)-1}\)=\(\frac{136}{136^2+136-1}\)

x=136,  A-B =\(\frac{x-1}{x^2-x-1}\)-\(\frac{x}{x^2+x-1}\) =\(\frac{x^3+x^2-x-x^2-x+1-x^3+x^2+x}{\left(x^2-1\right)^2-x^2}\)=\(\frac{x^2-x+2}{\left(x^2-1\right)^2-x^2}\)<0

=> A<B

b,A = \(\frac{456-333}{456}\)= 1-333/456       B=\(\frac{789-333}{789}\)= 1-333/789

=> A>B

c, 3/14<3/13<3/12<3/11<3/10 <2/5

M = 3/10+3/11+3/12+3/13+3/14 < 2/5 x5 = 2= N

Quy đồng tử số, ta có: \(\frac{1}{3}=\frac{1x5}{3x5}=\frac{5}{15}\)

Phân số nào có mẫu số nhỏ hơn thì lớn hơn

Mà: \(\frac{5}{15}>\frac{5}{123}\)

Vậy \(\frac{1}{3}>\frac{5}{123}\)

\(\frac{1}{3}\) và     \(\frac{5}{123}\)

Ta có :

\(\frac{1}{3}\) =     \(\frac{1x41}{3x41}\) = \(\frac{41}{123}\)         ;               \(\frac{5}{123}\) = Ta giữ nguyên phân số \(\frac{5}{123}\)

Ta có : \(\frac{41}{123}\) >      \(\frac{5}{123}\)

Vậy : \(\frac{1}{3}\)> \(\frac{5}{123}\)

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.

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

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

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

ta có:2/-7=-22/77

-3/11=-21/77

vì -22<-21 nên -22/77<-21/77

\(x=\frac{-2}{7}=\frac{-22}{77}\)

\(y=\frac{-3}{11}=\frac{-21}{77}\)

mà -22 < -21

=> x < y

\(\frac{2}{-7}\)= \(\frac{-2}{7}=\frac{-22}{77}\)

\(\frac{-3}{11}=\frac{-21}{77}\)

x < y vì -21 > -22

a, A=2015.2017=(2016-1)(2016+1)=2016²-1<2016²

Vậy A<B

Áp dụng \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\) (a;b;m \(\in\)N*)

Ta có:

\(A=\frac{3^{123}+1}{3^{125}+1}< \frac{3^{123}+1+2}{3^{125}+1+2}\)

\(A< \frac{3^{123}+3}{3^{125}+3}\)

\(A< \frac{3.\left(3^{122}+1\right)}{3.\left(3^{124}+1\right)}\)

\(A< \frac{3^{122}+1}{3^{124}+1}=B\)

=> A < B

\(9A=\frac{3^{125}+9}{3^{125}+1}=1+\frac{8}{3^{125}+1}\)

\(9B=\frac{3^{124}+9}{3^{124}+1}=1+\frac{8}{3^{124}+1}\)

Mà 3^125+1>3^124+1         =>\(\frac{8}{3^{125}+1}< \frac{8}{3^{124}+1}\)

Nên A<B

9A=\(\frac{3^{125}+9}{3^{125}+1}\)=\(1+\frac{8}{3^{125}+1}\)

9B=\(\frac{3^{124}+9}{3^{124}+1}\)=\(1+\frac{8}{3^{124}+1}\)

Vì \(\frac{8}{3^{125}+1}< \frac{8}{3^{124}+1}\)\(\Rightarrow9B>9A\)\(\Rightarrow B>A\)

Vậy B>A

a/

\(\frac{x}{z}=\frac{z}{y}\Rightarrow\frac{x^2}{z^2}=\frac{z^2}{y^2}=\frac{x^2+z^2}{z^2+y^2}\) (1)

Mà \(\frac{x}{z}=\frac{z}{y}\Rightarrow\frac{x^2}{z^2}=\frac{x}{z}.\frac{z}{y}=\frac{x}{y}\) (2)

Từ 91) và (2) \(\Rightarrow\frac{x^2+z^2}{y^2+z^2}=\frac{x}{y}\left(dpcm\right)\)

\(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)

\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)

\(\Rightarrow3^{223}>9^{111}>8^{111}>2^{332}\)

a) Từ \(\frac{x}{z}=\frac{z}{y}\)\(\Rightarrow\left(\frac{x}{z}\right)^2=\left(\frac{z}{y}\right)^2\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\left(\frac{x}{z}\right)^2=\left(\frac{z}{y}\right)^2=\frac{x^2}{z^2}=\frac{z^2}{y^2}=\frac{x^2+z^2}{y^2+z^2}\)

mà \(\left(\frac{x}{z}\right)^2=\frac{x}{z}.\frac{x}{z}=\frac{x}{z}.\frac{z}{y}=\frac{x}{y}\)

\(\Rightarrow\frac{x^2+z^2}{y^2+z^2}=\frac{x}{y}\)( đpcm )

b) Ta có: \(2^{332}< 2^{333}=2^{3.111}=\left(2^3\right)^{111}=8^{111}\)

              \(3^{223}>3^{222}=3^{2.111}=\left(3^2\right)^{111}=9^{111}\)

Vì \(8< 9\)\(\Rightarrow8^{111}< 9^{111}\)

\(\Rightarrow2^{332}< 8^{111}< 9^{111}< 3^{223}\)

\(\Rightarrow2^{332}< 3^{223}\)

\(\frac{x^2-y^2}{x^2+xy+y^2}=\frac{\left(x+y\right)\left(x-y\right)}{\left(x+y\right)^2-2xy}\left(1\right)\)

Vì \(x>y>0\) ta có :

\(\frac{x-y}{x+y}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}\left(2\right)\)

Do \(x>y>0\Leftrightarrow\left(x+y\right)^2-2xy< \left(x+y\right)^2\)\(\left(3\right)\)

Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow\frac{x-y}{x+y}< \frac{x^2-y^2}{x^2+xy+y^2}\)

Thanh Hằng Nguyễn copy bài à

Trong câu hỏi tương tự giải y hệt

Mình nghi lắm.

456 x 128 / 451 x 128 =58368/57728

123 x 451 / 128 x 451 = 55473/57728

so sánh : 58368/57728 ...>....  55473/ 57728 

vậy suy ra : 456/451 ....>.... 123/128 

tk mk nha mk nhanh nhất

\(\frac{456}{451}\) >    \(\frac{123}{128}\)tích cho mik nhé