Ta có: (2x-3y)*3=(x+2y)*2 => 6x-9y=2x+4y => 6x-2x=9y+4y => 4x=13y =>y/x=13/4

Câu 1:

\(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=x^2y^2+\frac{1}{256x^2y^2}+\frac{255}{256x^2y^2}+2\)

\(\ge\frac{1}{8}+2+\frac{255}{256x^2y^2}\)

Ta lại có: \(1=x+y\ge2\sqrt{xy}\Leftrightarrow1\ge16x^2y^2\)

\(\Rightarrow M\ge\frac{17}{8}+\frac{255}{16}=\frac{289}{16}\)

Dấu = xảy ra khi x=y=1/2

Áp dụng BDT Cauchy-Schwarz: \(\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)\ge\frac{1}{3x+3y+2z}\)

CMTT rồi cộng vế với vế ta có.\(VT\le\frac{1}{16}\cdot4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{3}{2}\)

Dấu = xảy ra khi x=y=z=1

Ta có:

\(\left(2x-3y\right).3=\left(x+2y\right).2\)

\(2x.3-3y.3=x.2+2y.2\)

\(6x-9y=2x+4y\)

\(6x-2x=9y+4y\)

\(4x=13y\)

Chia hai vế cho 4y, ta có:

\(\frac{4x}{4y}=\frac{13y}{4y}\)

\(=\frac{x}{y}=\frac{13}{4}\)

\(\Rightarrow\) Tỉ số giữa x và y là \(\frac{13}{4}\)

Ta có:

\(\frac{2x-3y}{x+2y}=\frac{2}{3}\)

<=> 3(2x-3y)= 2(x+2y)

<=> 6x - 9y = 2x + 4y

<=> 6x - 2x = 9y+4y

<=> 4x = 13y

=> \(\frac{x}{y}=\frac{13}{4}\)

khó nhỉ

ngu

\(\frac{2x-3y}{x+2y}=\frac{2}{3}=>\left(2x-3y\right).3=\left(x+2y\right).2=>6x-9y=2x+4y=>6x-2x=9y+4y\)

=>4x=13y

hay \(\frac{x}{y}=\frac{13}{4}\)

vậy gtri của tỉ số x/y là 13/4

2x² + 3y² = 5xy

=> 2x² + 3y² - 5xy = 0

=> 2 ( x² - 2xy + y² )  - xy + y² = 0

=> 2 ( x - y ) ² - y ( x - y ) = 0

=> ( x - y )[ 2( x - y ) - y ] = 0

=> ( x- y ) ( 2x - 2y - y ) = 0

=> ( x - y ) ( 2x - 3y ) = 0

TH1 : x - y = 0

=> x = y 

Thay x = y vào \(\frac{x+2y}{3x-y}\)

=> \(\frac{x+2y}{3x-y}=\frac{y+2y}{3y-y}\)\(=\frac{3y}{2y}=\frac{3}{2}\)

TH2 : 2x - 3y = 0

=> 2x = 3y

=> \(\frac{x}{y}=\frac{3}{2}\)

=> x = \(\frac{3}{2}.y\)

Thay x = \(\frac{3}{2}.y\)vào \(\frac{x+2y}{3x-y}\)

=> \(\frac{x+2y}{3x-y}=\frac{\frac{3}{2}.y+2y}{3.\frac{3}{2}y-y}\)\(=\frac{\frac{7}{2}.y}{\frac{7}{2}.y}=1\)

a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)

Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2

b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)

Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)

Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)