Question

[Nguyễn Văn Tiến]

Tính giá trị phân thức \(A=\frac{3x-2y}{3x+2y}\)

biết rằng \(9x^2+4y^2=20xy\)

và \(2y< 3x< 0.\)

Answer 1

[Cipher Thanh]

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