Question

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 giải giúp mình, mình cần gấp với ai giải đc mình tick cho

Answer 1

[HT.Phong (9A5)]

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k=>\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có:

\(VT=\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=\dfrac{\left(bd\right)\cdot k^2}{bd}=k^2\left(1\right)\)

\(VP=\dfrac{2009a^2+2010c^2}{2009b^2+2010d^2}=\dfrac{2009\cdot\left(bk\right)^2+2010\cdot\left(dk\right)^2}{2009b^2+2010d^2}\\ =\dfrac{2009b^2\cdot k^2+2010d^2\cdot k^2}{2009b^2+2010d^2}=\dfrac{k^2\cdot\left(2009b^2+2010d^2\right)}{2009b^2+2010d^2}\\ =k^2\left(2\right)\) 

Từ (1) và (2) => \(\dfrac{ac}{bd}=\dfrac{2009a^2+2010c^2}{2009b^2+2010d^2}\)