Từ \(9x^2+4y^2=20xy\Rightarrow9x^2-20xy+4y^2=0\)

\(\Leftrightarrow9x\left(x-2y\right)-2y\left(x-2y\right)=0\)\(\Leftrightarrow\left(x-2y\right)\left(9x-2y\right)=0\Leftrightarrow\orbr{\begin{cases}x=2y\\x=\frac{2}{9}y\end{cases}}\)

Với \(x=2y\Rightarrow A=\frac{3.2y+2y}{3.2y-2y}=\frac{8y}{4y}=2\)

Với \(x=\frac{2}{9}y\Rightarrow A=\frac{3.\frac{2}{9}y+2y}{3.\frac{2}{9}y-2y}=\frac{\frac{8}{3}y}{-\frac{4}{3}y}=-2\)

Từ \(9x^2+4y^2=20xy\Rightarrow9x^2-20xy+4y^2=0\)

\(\Leftrightarrow9x\left(x-2y\right)-2y\left(x-2y\right)=0\Leftrightarrow\left(x-2y\right)\left(9x-2y\right)=0\Leftrightarrow\orbr{\begin{cases}x=2y\\x=\frac{2}{9}y\end{cases}}\)

Với \(x=2y\Rightarrow A=\frac{3\cdot2y+2y}{3\cdot2y-2y}=\frac{8y}{4y}=2\)

Với \(x=\frac{2}{9}y\Rightarrow A=\frac{3\cdot\frac{2}{9}y+2y}{3\cdot\frac{2}{9}y-2y}=\frac{\frac{8}{3}y}{-\frac{4}{3}y}=-2\)

Ta có: \(A^2=\dfrac{\left(3x-2y\right)^2}{\left(3x+2y\right)^2}\)

\(=\dfrac{9x^2+4x^2-12xy}{9x^2+4x^2+12xy}\)

\(=\dfrac{20xy-12xy}{20x^2+12xy}\)

\(=\dfrac{8xy}{32xy}=\dfrac{1}{4}\)

\(\Leftrightarrow A\in\left\{\dfrac{1}{2};-\dfrac{1}{2}\right\}\)(1)

Vì 2y<3x<0 nên 3x-2y>0 và 3x+2y<0

hay \(A=\dfrac{3x-2y}{3x+2y}< 0\)(2)

Từ (1) và (2) suy ra \(A=-\dfrac{1}{2}\)

Vậy: \(A=-\dfrac{1}{2}\)

Ta có: \(9x^2+4y^2=20xy\Leftrightarrow9x^2-12xy+4y^2=8xy\Leftrightarrow\left(3x-2y\right)^2=8xy\) (1)

Mặt khác: \(9x^2+4y^2=20xy\Leftrightarrow9x^2+12xy+4y^2=32xy\Leftrightarrow\left(3x+2y\right)^2=32xy\) (2)

Từ (1) và (2) => \(\frac{\left(3x-2y\right)^2}{\left(3x+2y\right)^2}=\frac{8xy}{32xy}\Leftrightarrow\left(\frac{3x-2y}{3x+2y}\right)^2=\frac{1}{4}\Leftrightarrow\frac{3x-2y}{3x+2y}=\pm\frac{1}{2}\)

Mà \(2y< 3x< 0\Rightarrow A=\frac{3x-2y}{3x+2y}=\frac{-1}{2}\)

Ta có: \(A^2=\frac{9x^2+4y^2-12xy}{9x^2+4y^2+12xy}=\frac{20xy-12xy}{20xy+12xy}=\frac{8xy}{32xy}=\frac{1}{4}\)

Vì \(2y< 3x< 0\Rightarrow3x-2y>0,3x+2y< 0\Rightarrow A< 0\)

Vậy A= \(\frac{-1}{2}\)

Ta có :

\(A^2=\frac{9x^2+4y^2-12xy}{9x^2+4y^2+12xy}\)\(=\frac{20xy-12xy}{20xy+12xy}\)\(=\frac{8xy}{32xy}\)\(=\frac{1}{4}\)

\(Do\)\(2y< 3x< 0\Rightarrow3x-2y>0;3x+2y< 0\Rightarrow A< 0\)

Vậy \(A=-\frac{1}{2}\)

Ta có \(9x^2+4y^2=20xy\Leftrightarrow9x^2+2.3x.2y+4y^2=8xy\Leftrightarrow\left(3x+2y\right)^2=8xy\)\(32xy\)

Mặt khác \(9x^2+4y^2=20xy\Leftrightarrow9x^2-2.3x.2y+4y^2=8xy\Leftrightarrow\left(3x-2y\right)^2=8xy\)

\(\Rightarrow\frac{\left(3x-2y\right)^2}{\left(3x+2y\right)^2}=\frac{8xy}{32xy}=\frac{1}{4}\)\(\Leftrightarrow\left(\frac{3x-2y}{3x+2y}\right)^2=\frac{1}{4}\Leftrightarrow\frac{3x-2y}{3x+2y}=+-\frac{1}{2}\)

Do \(2y< 3x< 0\Rightarrow A=-\frac{1}{2}\)

ta có 

9x²+12xy+4y²=32xy

=>(3x+2y)²=32xy =>3x+2y=\(\sqrt{32xy}\)

mặt khác

9x²-12xy+4y²=8xy

=>(3x-2y)²=8xy  =>3x-2y=\(\sqrt{8xy}\)

vậy \(\frac{3x-2y}{3x+2y}=\frac{\sqrt{8xy}}{\sqrt{32xy}}\)

=0,5

đề này có trong violimpic vòng 15

hôm qua mình đi thi có gặp bài này ko bt sai hay đúng nữa

mà hình như mình làm sai dấu

Ta có : b,  \((3x-2y)^2=9x^2-12xy+4y^2=20xy-12xy=8xy\)

\(\Rightarrow3x-2y=\sqrt{8xy}\)                             \((1)\)

\((3x+2y)^2=9x^2+12xy+4y^2=20xy+12xy=32xy\)

\(\Rightarrow3x+2y=\sqrt{32xy}\)                             \((2)\)

Từ \((1)\) và      \((2)\), suy ra :

\(\Rightarrow\frac{3x-2y}{3x+2y}=\frac{\sqrt{8xy}}{\sqrt{32xy}}=0,5\)