\(9x^2+4y^2=20xy\)
\(\Leftrightarrow\left(3x\right)^2+\left(2y\right)^2-20xy=0\)
\(\Leftrightarrow\left(3x\right)^2+\left(2y\right)^2-12xy-8xy=0\)
\(\Leftrightarrow\left(3x-2y\right)^2-8xy=0\)
\(\Leftrightarrow\left(3x-2y\right)^2=8xy\)
\(\Leftrightarrow3x-2y=\sqrt{8xy}\)(1)
- \(9x^2+4y^2=20xy\)
\(\Leftrightarrow\left(3x\right)^2+\left(2y\right)^2-20xy=0\)
\(\Leftrightarrow\left(3x\right)^2+\left(2y\right)^2+12xy-32xy=0\)
\(\Leftrightarrow\left(3x+2y\right)^2-32xy=0\)
\(\Leftrightarrow\left(3x+2y\right)^2=32xy\)
\(\Leftrightarrow3x+2y=\sqrt{32xy}\)(2)
Thay (1) và (2) vào A, ta có:
\(A=\dfrac{3x-2y}{3x+2y}=\dfrac{\sqrt{8xy}}{\sqrt{32xy}}=0,5\)
\(9x^2+4y^2=20xy\)
=>\(\left(9x^2-18xy\right)-\left(2xy-4y^2\right)=0\)
=>\(9x\left(x-2y\right)-2y\left(x-2y\right)=0\)
=>\(\left(x-2y\right)\left(9x-2y\right)=0\)
=>\(x-2y=0\) (vì 2y<3x<0 nên 9x\(\ne2y\) )
=>x=2y
Thay x=2y vào A ta có
\(A=\dfrac{6y-2y}{6y+2y}=\dfrac{4y}{8y}=0,5\)
Vậy A=0,5
\(9x^2+4y^2=20xy\Rightarrow\left\{{}\begin{matrix}\left[\left(3x\right)^2-2.\left(3x\right).\left(2y\right)+\left(2y\right)^2\right]=8xy\left(1\right)\\\left[\left(3x\right)^2+2.\left(3x\right).\left(2y\right)+\left(2y\right)^2\right]=32xy\left(2\right)\end{matrix}\right.\)Lấy (1) chia cho (2)
\(\left|\dfrac{3x-2y}{3x+2y}\right|=\left|A\right|=\dfrac{8}{32}=\dfrac{1}{2}\)
\(2y< 3x< 0\Rightarrow\left\{{}\begin{matrix}3x-2y>0\\3x+2y< 0\end{matrix}\right.\)
\(A< 0\Rightarrow A=-\dfrac{1}{2}\)
