Question

Cho 4a² + b² = 5ab và 2a > b > 0. Tính P = ab/ ( 4a² - b²)

Answer 1

ta có: \(4a^2+b^2=5ab< =>4a^2-5ab+b^2=0< =>4a^2-4ab-ab+b^2=0< =>4a\left(a-b\right)-b\left(a-b\right)=0< =>\left(a-b\right)\left(4a-b\right)=0\)

do 2a>b>0=>4a>b>0=> 4a-b khác 0

=> a-b=0<=>a=b

P=\(\dfrac{ab}{4a^2-b^2}=\dfrac{ab}{\left(2a-b\right)\left(2a+b\right)}=\dfrac{ab}{\left(2a-a\right)\left(2a+a\right)}=\dfrac{a^2}{3a^2}=\dfrac{1}{3}\)

vậy............

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