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 \(A=-\frac{1}{2}\)

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

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: 9x^2+4y^2=20xy=> 9x^2-2.2.3xy+4y^2=8xy

=> (3x-2y)^2=8xy

mặt khác 9x^2+4y^2=20xy=> 9x^2+2.2.3xy+4y^2=32xy

=>(3x+2y)^2=32xy

=>(3x-2y)^2/(3x+2y)^2=8xy/32xy=1/4

=>(3x-2y)/(3x+2y)=căn 1/4=1/2 hoặc -1/2

mà x<2y=>x=-1/2

Ta có:

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

\(\Leftrightarrow9x^2-20xy+4y^2=0\)

\(\Leftrightarrow9x^2-18xy-2xy+4y^2=0\)

\(\Leftrightarrow9x\left(x-2y\right)-2y\left(x-2y\right)=0\)

\(\Leftrightarrow\left(x-2y\right)\left(9x-2y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2y\\9x=2y\end{matrix}\right.\)

Mà \(x< 2y\) nên \(9x=2y\Leftrightarrow x=\dfrac{2}{9}y\) (1)

Thay (1) vào A ta được:

\(A=\dfrac{3.\dfrac{2}{9}y-2y}{3.\dfrac{2}{9}y+2y}=\dfrac{y\left(\dfrac{2}{3}-2\right)}{y\left(\dfrac{2}{3}+2\right)}=\dfrac{-\dfrac{4}{3}}{\dfrac{8}{3}}=-\dfrac{1}{2}\)

Vậy..................................

a) ĐKXĐ : \(x+y\ne0\)

\(x^2-2y^2=xy\)

\(x^2-y^2-y^2-xy=0\)

\(\left(x-y\right)\left(x+y\right)-y\left(y+x\right)=0\)

\(\left(x+y\right)\left(x-2y\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+y=0\left(Loai\right)\\x-2y=0\left(Chon\right)\end{matrix}\right.\)

Với x - 2y = 0 ta có x = 2y

Thay x = 2y vào A ta có :

\(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)

a)

Ta có:

\(x^2-2y^2=xy\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-y\left(y+x\right)=0\)\(\Leftrightarrow\left(x+y\right)\left(x-y-y\right)=\left(x+y\right)\left(x-2y\right)=0\)

=>x-2y=0=>x=2y

Thế vào A rùi giải