Ta có:
\(P=\dfrac{5x-4y}{5x+4y}\)
\(\Leftrightarrow P^2=\left(\dfrac{5x-4y}{5x+4y}\right)^2\)
\(\Leftrightarrow P^2=\dfrac{\left(5x-4y\right)^2}{\left(5x+4y\right)^2}\)
\(\Leftrightarrow P^2=\dfrac{\left(5x\right)^2-2\cdot5x\cdot4y+\left(4y\right)^2}{\left(5x\right)^2+2\cdot5x\cdot4y+\left(4y\right)^2}\)
\(\Leftrightarrow P^2=\dfrac{\left(25x^2+16y^2\right)-40xy}{\left(25x^2+16y^2\right)+40xy}\)
Thay \(25x^2+16y^2=50xy\) vào ta có:
\(P^2=\dfrac{50xy-40xy}{50xy+40xy}=\dfrac{10xy}{90xy}=\dfrac{1}{9}=\left(\dfrac{1}{3}\right)^2\)
Mà: \(4y< 5x< 0\)
Nên: \(P=\dfrac{5x-4y}{5x+4y}< 0\)
Vậy: \(P=-\dfrac{1}{3}\)
25x^2+16y^2=50xy
=>25x^2-50xy+16y^2=0
=>25x^2-10xy-40xy+16y^2=0
=>5x(5x-2y)-8y(5x-2y)=0
=>(5x-2y)(5x-8y)=0
=>5x=2y hoặc 5x=8y
5x>4y
=>5x=8y
=>x/8=y/5=k
=>x=8k; y=5k
\(P=\dfrac{5\cdot8k-4\cdot5k}{5\cdot8k+4\cdot5k}=\dfrac{40-20}{40+20}=\dfrac{1}{3}\)
