Question

Tính

\(\frac{\text{ }\left[\left(e-m\right)^2-\left(e+m\right)^2\right]\text{ }\left[\left(y-1\right)^2-\left(y+1\right)^2\right]}{a.16.n.h}\)\(.\frac{e}{u^{-1}}\)

Answer 1

mình giải theo cách lớp 8 nha!!!

\(\frac{\left[\left(e-m\right)^2-\left(e+m\right)^2\right]\left[\left(y-1\right)^2-\left(y+1\right)^2\right]}{16ahn}\cdot\frac{e}{u^{-1}}\)

\(=\frac{\left(e-m-e-m\right)\left(e-m+e+m\right)\left(y-1-y-1\right)\left(y-1+y+1\right)}{16ahn}\cdot eu\)

\(=\frac{\left(-2m\right)\left(2e\right)\left(-2\right)\left(2y\right)}{16ahn}\cdot eu\)

\(=\frac{16mey}{16ahn}\cdot eu\)

\(=\frac{e^2myu}{ahn}\)