Question

Biết 9x² + 4y² = 20xy và 2y < 3x < 0. Khi đó giá trị của là A = ...........

Thanks~~ biết ơn nhìu lắm ạ!

Answer 1

Theo bài ra , ta có :

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

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

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

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

\(\Leftrightarrow\left[\begin{matrix}x-2y=0\\9x-2y=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=2y\\9x=2y\end{matrix}\right.\)

Thay x = 2y vào A ta đk :

\(A=\frac{3x-2y}{3x+2y}=\frac{3.2y-2y}{3.2y+2y}=\frac{4y}{8y}=\frac{4}{8}=\frac{1}{2}\)

Vậy \(A=\frac{1}{2}\)

Chúc bạn học tốt =))

Answer 2

Ta có: A=\(\frac{3x-2y}{3x+2y}\)

=>A²=\(\frac{\left(3x-2y\right)^2}{\left(3x+2y\right)^2}\)=\(\frac{9x^2-12xy+4y^2}{9x^2+12xy+4y^2}\)=\(\frac{\left(9x^2+4y^2\right)-12xy}{\left(9x^2+4y^2\right)+12xy}\)=\(\frac{20xy-12xy}{20xy+12xy}\)=\(\frac{8xy}{32xy}\)=\(\frac{1}{4}\)

^{=>\(\left\{\begin{matrix}A=\frac{1}{2}\\A=\frac{-1}{2}\end{matrix}\right.\)}

^{Do 2y<3x<0}

^{=>\(\frac{3x-2y}{3x+2y}\)}<0

=>A=\(\frac{-1}{2}\)

Answer 3

-0,5