Question

Biết 2x > y > 0 và 4x² + y² = 5xy. Tính M = \(\frac{xy}{4x^2-y^2}\)

Answer 1

Từ gt \(4x^2+y^2=5xy\)

\(\Leftrightarrow4x^2-4xy+y^2-xy=0\)

\(\Leftrightarrow4x\left(x-y\right)+y\left(y-x\right)=0\)

\(\Leftrightarrow4x\left(x-y\right)-y\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(4x-y\right)=0\)

Vì \(2x>y>0\Rightarrow4x>y\Leftrightarrow4x-y>0\)

\(\Rightarrow x-y=0\Leftrightarrow x=y\)

Thay vào M:

\(M=\frac{xy}{4x^2-y^2}=\frac{x^2}{4x^2-x^2}=\frac{x^2}{3x^2}=\frac{1}{3}\)

Answer 2

ta có :

4x²+y²=5xy

⇔ 4x²+y²-5xy=0

⇔ 4x² - 4xy + y²-xy=0

⇔4x(x-y) - y(x-y) = 0

⇔ (x - y)(4x-y)=0

vì 2x > y > 0 nên 4x-y>0

⇒ x-y=0 ⇒ x = y

⇒M= \(\frac{xy}{4x^2-y^2}\)=\(\frac{x^2}{4x^2-x^2}=\frac{x^2}{3x^2}=\frac{1}{3}\)

vậy M = \(\frac{1}{3}\)