= \(\dfrac{1}{2}\)nha

\(\dfrac{3x-2y}{3x+2y}=\dfrac{\left(3x-2y\right)^2}{\left(3x+2y\right)^2}=\dfrac{9x^2+4y^2-12xy}{9x^2+4y^2+12xy}=\dfrac{1}{4}\)

thay từ đề vào ok

\(\left\{{}\begin{matrix}2y< 3x< 0\\\left\{{}\begin{matrix}3x-2y>0\\3x+2y< 0\end{matrix}\right.\end{matrix}\right.\) \(\left\{{}\begin{matrix}A^2=\dfrac{12xy}{32xy}=\dfrac{12}{32}=\dfrac{3}{8}\\A< 0\Rightarrow A=-\sqrt{\dfrac{3}{8}}\end{matrix}\right.\)

\(A=\dfrac{2x^2\left(3x-4y+2\right)}{x\left(3x+y\right)\left(3x-y\right)}=\dfrac{2x\left(3x-4y+2\right)}{\left(3x+y\right)\left(3x-y\right)}\\ A=\dfrac{2\left(3-8+2\right)}{\left(3+2\right)\left(3-2\right)}=\dfrac{2\left(-3\right)}{5}=\dfrac{-6}{5}\)

\(\left|x\right|=1\Rightarrow x=\pm1;\left|y\right|=3\Rightarrow y=\pm3\)

Với x = 1, y = 3 => A = 3.1² - 3.1².3 + 2.3² = 3 - 9 + 18 = 12

Với x = -1, y = -3 => A = 3.(-1)² - 3.(-1)².(-3) + 2.(-3)² = 3 + 9 + 18 = 30

Với x = 1, y = -3 => A = 3.1² - 3.1².(-3) + 2.(-3)² = 3 + 9 + 18 = 30

Với x = -1, y = 3 => A = 3.(-1)² - 3.(-1)².3 + 2.3² = 3 - 9 + 18 = 12

Vậy...

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

ĐK: \(3x\ne\pm y;x\ne0\)

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

= \(\dfrac{3x\left(3x-y\right)-x\left(3x+y\right)+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{6x^2-4xy+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{2x\left(3x-2y+1\right)}{\left(3x-y\right)\left(3x+y\right)}\)

Thay x = 1; y=2, ta có:

A = \(\dfrac{2.1\left(3.1-2.2+1\right)}{\left(3.1-2\right)\left(3.1+2\right)}=0\)