L
lan_anh_a


[tex] \frac{[(e - m)^2 – (e + m)^2].[(y-1)^2 - (y + 1)^2] }{a16nh} . \frac{e}{u^{-1}}[/tex]
kachia said:Tạm hiểu :
[tex]\huge P= \frac{[(e-m)^2-(e+m)^2][(y-1)^2-(y+1)^2]}{a16nh}.\frac{e}{u^{-1}}[/tex]
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